Erdös numbers

Micheal Pawliuk — Wed, 16 May 2012 19:04:05 +0000

(This talk was given as part of the What We Talk About lecture series at No One Writes to the Colonel on May 17, 2012)

Paul Erdös

Math is many things: beautiful, fundamental, universal, but ultimately, math is hard. So hard in fact that mathematicians often need to phone up their other mathematician friends for help. The idea of the crazy-maned recluse furiously working in his office alone is an outdated one. While some modern mathematicians still fit this bill, collaboration is increasingly the norm. By the year 2000, the number of mathematics papers with a single author had shrunk to 50%. More and more people are tackling difficult math problems as a team.

No one embodies the idea of mathematical collaboration more than Paul Erdös (pronounced err-desh or air-dish), a Hungarian mathematician who lived in the twentieth century. A legendary figure in mathematics, Erdös published around 1500 papers and had around 500 co-authors. To contrast, most mathematicians write 7 papers in their entire life! Erdös was heavily in support of working together to solve math problems and questions, and also had incredible mathematical taste. He asked very interesting questions and would often attach a dollar amount to the questions. If you were clever enough to solve one of these Erdös questions, Paul Erdös himself would send you a cheque. These cheques were so revered in mathematics that often people frame them rather than cash them. More information on his very interesting life is available here.

It is in this context that the idea of the Erdös number emerged. In the 60s, one of Erdös’ friends presented the idea of the Erdös number to measure how close you were to Erdös in terms of collaboration. So Erdös himself has Erdös number 0 and all of his 511 direct co-authors have Erdös number 1. Now if you co-author a paper with say, Terrence Tao who has Erdös number 1, you would have Erdös number 2! If, like me, you have never published a paper, or you have never co-authored a paper, you would have an infinite Erdös number.

I would say that most mathematicians know their Erdös number (or can easily find it out); it is kind of like a feather in their hat. Your Erdös number on its own won’t get you a job, a diploma or a date, but it does look nice. Some people have been known to write papers with collaborators simply to lower their Erdös number (I think someone was selling the opportunity to co-author with them on Ebay) but this is relatively rare.

Now for some facts (all of which I found on the helpful Erdös Number Project). The smallest Erdös number is of course 0, but the largest is 13 (or 15 depending on what you consider to be a paper). Arguably the most famous mathematician, Andrew Wiles who solved Fermat’s Last Theorem, has Erdös number 3. Looking at the who’s who of brilliant mathematicians we see that every Fields medalist (kind of the math Nobel prize) has had Erdös number 5 or less, with most having number 2 or 3. In contrast the upper bound for the Nobel Prize in Medicine is 11.

Many non-mathematicians have Erdös numbers as well. Bill Gates, Stephen Hawking and Carl Sagan each have number 4. Some fields are even known for having low Erdös numbers. Biology and Linguistics are examples, as in Linguistics, Noam Chomsky has Erdös number 4.

Stephen Hawking on The Simpsons

Even more surprisingly, there are many actors with Erdös numbers. Natalie Portman (5) and Colin Firth (6) are counted among the various co-co-co-co-collaborators of Erdös. This leads to the inevitable connection with the “6 degrees of Kevin Bacon” game. You remember the game where I give you an actor and you try to connect them to Kevin Bacon by saying “A was in a film with B who was in a film with C … who was in a film with Kevin Bacon”.

Thus you have an actor’s “Bacon number”, the length of the shortest path of that sort to Kevin Bacon. Going even further you have “Erdös-Bacon numbers” (it’s a real thing, look it up). This is the sum of a person’s Erdös number and their Bacon number. Of the people I mentioned, Carl Sagan has EB number 7, Stephen Hawking has EB number 7 (if you count his appearance on the Simpsons), Colin Firth has EB number 7 and Natalie Portman has EB number 6. The smallest (non-disputed) EB number is 3 which belongs to Bruce Reznick a math professor with Erdös number 1 that appeared as an extra in Pretty Maids all in a Row, which gives him Bacon number 2.

By the way, this is from the Wikipedia plot synopsis of Pretty Maids All in a Row: “The story is set in Oceanfront High School, a fictitious American high school in the height of the sexual revolution. Young female students are being targeted by an unknown serial killer. Meanwhile, a male student called Ponce is experiencing sexual frustration, surrounded by a seemingly unending stream of beautiful and sexually provocative classmates.”

To end, let’s examine the famous baseball player Hank Aaron (who in my mind is still the all time leader in home runs) who has Bacon number 2 (he was in “Summer Catch” with Susan Gardner who was in “In the Cut” with Kevin Bacon). Hank Aaron has also signed the same baseball as Paul Erdös, giving him a tenuous Erdös number 1. So you could say that Hank Aaron also has an Erdös-Bacon number of 3!

Hank Aaron

Envelope forcing

François G. Dorais — Sun, 06 May 2012 18:02:01 +0000

$\newcommand{\RCA}{\mathsf{RCA}_0}\newcommand{\Ind}[1]{\mathsf{I}{#1}}\newcommand{\RT}[2]{\mathsf{RT}^{#1}_{#2}}\newcommand{\Wk}{\mathsf{W}}\newcommand{\HWk}{\mathsf{HW}}\newcommand{\Pth}{\mathsf{Path}}\newcommand{\St}{\mathsf{S}}\newcommand{\Mix}{\mathsf{Mixed}}\newcommand{\PP}{\mathbb{P}}\newcommand{\MN}{\mathcal{N}}$ In a recent paper [1], Jared Corduan and I considered various notions of combinatorial indecomposability for finite ordinal powers of $\omega.$ In this process, we uncovered two weak forms of Ramsey’s theorem for pairs: The Weak Ramsey Theorem ($\Wk\RT2k$). For every coloring $c:\N^2\to\set{0,\dots,k-1}$ there are a color $d \lt k$ and an infinite set [...]

The chromatic numbers of the Erdos-Hajnal graphs

Assaf Rinot — Sat, 05 May 2012 03:47:58 +0000

Recall that a coloring $c:G\rightarrow\kappa$ of an (undirected) graph $(G,E)$ is said to be chromatic if $c(v_1)\neq c(v_2)$ whenever $\{v_1,v_2\}\in E$. Then, the chromatic number of a graph $(G,E)$ is the least cardinal $\kappa$ for which there exists a chromatic coloring $c:G\rightarrow\kappa$.
Motivated by a paper I found yesterday in the arXiv, this post will be dedicated to exemplifying incompactness properties of the chromatic number measure. More specifically, we shall give a construction of Shelah, of a graph of size $\lambda$ such that all of its subgraphs of size $<\lambda$ are countably chromatic, while the graph itself is not.

Before stating and proving Shelah’s theorem, we mention a somewhat simpler (yet, canonical) test-case. To every cardinal $\lambda$, one associates the Erdos-Hajnal graph $EH(\lambda)=({}^\lambda\omega,\bot)$, where $f\bot g$ stands for the set $\{ \alpha<\lambda\mid f(\alpha)=g(\alpha)\}$ being bounded in $\lambda$. Here is a list of fairly simple observations:

if $\lambda$ is a regular cardinal, then any subgraph of $EH(\lambda)$ of size $<\lambda$ is countably chromatic;
the chromatic number of $EH(\aleph_0)$ equals $2^{\aleph_0}$;
the chromatic number of $EH(\aleph_1)$ is bigger than $\aleph_1$;
if $(G,E)$ is a graph of size $\lambda$ such that every subgraph of size $<\lambda$ is countably chromatic, then the chromatic number of $EH(\lambda)$ is $\ge$ the chromatic number of $(G,E)$. This shows that $EH(\lambda)$ is a canonical witness to incompactness of this sort, whenever exists.
why did I write “whenever exists” in the previous item? well, for instance, if $\lambda$ is a measurable cardinal, then every function $f:\lambda\rightarrow\omega$ is fixed on a measure one set, and hence $EH(\text{measurable})$ is countably chromatic!

Now, on to the theorem.

Theorem (Shelah, 2012). Suppose that $\lambda$ is a regular cardinal $>\aleph_1$.
If $E^{\lambda}_\omega$ admits a non-reflecting stationary subset, and $\mu^{\aleph_0}\le\lambda$ for all $\mu<\lambda$, then there exists a graph $(G,E)$ of size $\lambda$ such that any of its subgraphs of size $<\lambda$ is countably chromatic, while $(G,E)$ is not countably chromatic.
In particular, the above assumptions entails that $EH(\lambda)$ is not countably chromatic.
Proof. Fix a regressive surjection $f:\lambda\rightarrow\lambda$ such that the preimage of every singleton is cofinal in $\lambda$. Denote $F_\delta:=\{\gamma<\lambda\mid f(\gamma)=\delta\}$. Let $S\subseteq E^{\lambda}_\omega$ be a non-reflecting stationary subset, and put $T:=f^{-1}[S]$.
As the underlying set of our graph, we take: $$G:=\{ (\alpha,\beta)\in \lambda\times T\mid \alpha<f(\beta)\}.$$

Next, we fix a club guessing sequence $\langle C_\delta\mid\delta\in S\rangle$. For all $\delta\in S$, let $\{ \delta_n\mid n<\omega\}$ denote the increasing enumeration of $C_\delta$. Let $\Gamma_\delta$ denote the collection of all sequences $\langle \beta_n\mid n<\omega\rangle$ that satisfies for all $n<\omega$:

$(\beta_{2n},\beta_{2n+1})\in G$;
$\delta_n<\beta_{2n}<\beta_{2n+1}<\delta_{n+1}$.

As $|\delta|^{\aleph_0}\le\lambda=|F_\delta|$, we may fix a surjection $g_\delta:F_\delta\rightarrow\Gamma_\delta$.
Finally, the set of edges of the constructed graph will be:
$$E:=\left\{\{(\beta_{2m},\beta_{2m+1}),(\delta_0,\gamma)\}\mid m<\omega, \delta\in S, \gamma\in F_\delta, g_\delta(\gamma)=\langle\beta_n\mid n<\omega\rangle\right\}.$$

Subclaim. $(G,E)$ is not countably chromatic.
Proof. Suppose not, as witnessed by some coloring $c:G\rightarrow\omega$. For each $\tau<\lambda$, let $u_\tau:=\{ c(\alpha,\beta)\mid (\alpha,\beta)\in G, \alpha\ge\tau\}$. Then $\{ u_\tau\mid \tau<\lambda\}$ is a descending chain, and since $\text{cf}(\lambda)>\omega$, the chain must stabilize at some $\tau^*<\lambda$. Denote $u^*:=u_{\tau^*}$.
Consider the club$$D:=\{ \delta<\lambda\mid \forall\tau<\delta\forall n\in u^*\exists(\alpha,\beta)\in G\text{ with }\tau<\alpha<\beta<\delta\text{ s.t. }c(\alpha,\beta)=n\}.$$
Pick $\delta\in S$ such that $C_\delta\subseteq D\setminus\tau^*$.
It follows that we may find a sequence $\langle \beta_n\mid n<\omega\rangle$ such that for all $n<\omega$:

$(\beta_{2n},\beta_{2n+1})\in G$;
$\delta_n<\beta_{2n}<\beta_{2n+1}<\delta_{n+1}$;
if $n\in u^*$, then $c(\beta_{2n},\beta_{2n+1})=n$.

Pick $\gamma\in F_\delta$ such that $g_\delta(\gamma)=\langle \beta_n\mid n<\omega\rangle$. Then for all $n<\omega$, we have $\{(\beta_{2n},\beta_{2n+1}),(\delta_0,\gamma)\}\in E$, and hence $c(\delta_0,\gamma)\neq c(\beta_{2n},\beta_{2n+1})$. In particular, $$c(\delta_0,\gamma)\in\omega\setminus u^*.$$
On the other hand, $\tau^*\le\delta_0<\gamma$, and so $c(\delta_0,\gamma)\in u_{\tau^*}=u^*$. This is a contradiction. $\blacksquare$

Write $G=\bigcup_{i<\lambda}G_i$, where $G_i:=\{ (\alpha,\beta)\in G \mid f(\beta)<i\}$.

Note. if $i\not\in S$, then $G_{i+1}=G_i$.
Proof. If $(\alpha,\beta)\in G_{i+1}$, then $\beta\in T$ and $f(\beta)<i+1$. As $f(\beta)\in S$, while $i\not\in S$, we infer that $f(\beta)<i$, and so $(\alpha,\beta)\in G_i$. $\blacksquare$

Main subclaim. Suppose that $u$ is an infinite subset of $\omega$.
Then, for every $i\le j<\lambda$ with $i\not\in S$, and a chromatic coloring $c:G_i\rightarrow\omega$, there exists a chromatic coloring $c’:G_j\rightarrow\omega$ such that $c’\restriction G_i=c$, and $c’[G_j\setminus G_i]\subseteq u$.
Proof. Given $u\in[\omega]^\omega$, let $\{ u_n\mid n<\omega\}$ be some partition of $u$ into mutually-disjoint infinite sets. We now prove the claim by induction on $j<\lambda$.

$\blacktriangleright$ The case $j=0$ is trivial. $\blacktriangleleft$

$\blacktriangleright$ If $j$ is a limit non-zero ordinal, let us pick a club $e$ in $j$ such that $\min(e)=i$. As $S$ is non-reflecting, we may moreover assume that $e\cap S=\emptyset$.
Now, build an ascending chain of chromatic colorings $\{ c_k:G_k\rightarrow\omega\mid k\in e\}$ by induction on $k\in e$:

for $k=\min(e)$, let $c_k:=c$;
for $k\in\text{nacc}(e)$, let $i’:=\sup(e\cap k)$, and $j’:=k$. Then $i’<j’<j$, and $i’\not\in S$, so by the induction hypothesis, we may pick a chromatic coloring $c_k:G_k\rightarrow\omega$ that extends $c_{i’}$, and such that $c_k[G_k\setminus G_{i'}]\subseteq u$.
for $k\in\text{acc}(e)$, let $c_k:=\bigcup_{i’<k}c_{i’}$. Then $c_k$ is chromatic as the limit of chromatic colorings.

Finally, put $c’:=\bigcup_{k<j}c_k$. Then $c’$ has all the desired properties. $\blacktriangleleft$

$\blacktriangleright$ If $j$ is a successor ordinal, and $j-1\not\in S$, then $G_j=G_{j-1}$, and so we may appeal to the induction hypothesis for $j-1$. $\blacktriangleleft$

$\blacktriangleright$ If $j$ is a successor ordinal, and $j-1\in S$, let us denote $\delta:=j-1$. Since $i\le\delta$ and $i\not\in S$, we get that $i<\delta$, so let us fix a large enough $n^*<\omega$ so that $\delta_{n^*}\ge i$.
Let

$j(n):=\delta_{(n^*+n)}+1$ for all $n<\omega$;
$i(0):=i$, and $i({n+1}):=j(n)$ for all $n<\omega$.

Note that for all $n<\omega$, we have $i\le i(n)<j(n)<j$ and $i(n)\not\in S$.
So it follows from the induction hypothesis, that there exists an increasing chain of chromatic colorings $\{ c_n:G_{j(n)}\rightarrow\omega\mid n<\omega\}$ that extends $c$, and such that $c_{n}[G_{j(n)}\setminus G_{i(n)}]\subseteq u_{n}$, for all $n<\omega$. Note that $\text{dom}(\bigcup_{n<\omega}c_n)=G_\delta$.
Finally, let $c’:G_j\rightarrow \omega$ be some coloring that satisfies:

$c’\restriction G_\delta=\bigcup_{n<\omega}c_n$;
for every $\gamma\in F_\delta$, if $g_\delta(\gamma)=\langle \beta_n\mid n<\omega\rangle$, then $$c’(\delta_0,\gamma)\in u_0\setminus\{ c(\beta_{2m},\beta_{2m+1})\mid m<n^*\}.$$

We need to verify that $c’$ is chromatic.
For this, fix $x,y\in G_j$ such that $\{x,y\}\in E$, and let us verify that $c’(x)\neq c’(y)$.
Of course, we may assume that $\{ x,y\}\setminus G_\delta\not=\emptyset$, because otherwise $\{ x,y\}\subseteq\text{dom}(c_n)$ for some $n<\omega$.

Next, by the definition of $E$, there exists $\delta^*\in S$, $m<\omega$, and $\gamma<\lambda$ such that:

$f(\gamma)=\delta^*$;
$g_{\delta^*}(\gamma)=\langle\beta_n\mid n<\omega\rangle$;
$\{ x,y\}=\{ (\beta_{2m},\beta_{2m+1}) , (\delta^*_0,\gamma) \}$.

As $(\beta_{2m},\beta_{2m+1})\in G_j$, we know that $f(\beta_{2m+1})\le\delta$. In addition, $f(\beta_{2m+1})\neq\delta$, because otherwise $j=\delta+1=f(\beta_{2m+1})+1\le\beta_{2m+1}<\delta^*_{m+1}<\delta^*=f(\gamma)$, contradicting the fact that $(\delta^*_0,\gamma)\in G_j$.

So $f(\beta_{2m+1})<\delta$, that is $(\beta_{2m},\beta_{2m+1})\in G_\delta$, and since $\{x,y\}\setminus G_\delta\neq\emptyset$, we conclude that $f(\gamma)=\delta$, and $(\delta^*_0,\gamma)=(\delta_0,\gamma)$.

Now, if $m<n^*$, then by the very choice of $c’$, we have $$c’(\delta_0,\gamma)\in u_0\setminus \{c(\beta_{2m},\beta_{2m+1})\}.$$

If $m\ge n^*$, then $n:=m-n^*$ is a natural number, and so $$\delta_m<\beta_{2m}<\beta_{2m+1}<\delta_{m+1}$$ entails
$$i(n+1)=j(n)=\delta_m+1\le\beta_{2m}<\beta_{2m+1}<\delta_{m+1}<j(n+1).$$
Since $(\beta_{2m},\beta_{2m+1})\in G$, we get that $i(n+1)\le \beta_{2m}<f(\beta_{2m+1})$. Since $f$ is regressive, we get that $f(\beta_{2m+1})<\beta_{2m+1}<j(n+1)$. Altogether, $$(\beta_{2m},\beta_{2m+1})\in G_{j(n+1)}\setminus G_{i(n+1)},$$ and hence $$c’(\beta_{2m},\beta_{2m+1})\in u_{n+1}.$$
In particular, $$c’(\beta_{2m},\beta_{2m+1})\not\in u_0.$$This concludes the last case. $\blacktriangleleft$
This completes the proof of the main submclaim. $\blacksquare$

It follows from the main subclaim that for every $i<\lambda$, the subgraph $(G_i,E\restriction [G_i]^2)$ is countably chromatic. As $\lambda$ is regular, this in particular entails that every subgraph of $(G,E)$ of size $<\lambda$ is countably chromatic. $\blacksquare$

Corollary. If $\lambda$ is a strong limit singular cardinal, and $EH(\lambda^+)$ is countably chromatic, then $0^\sharp$ exists.
Proof. The covering lemma implies $\lambda^{\aleph_0}\le\lambda^+$ and $\square_\lambda$ for every strong limit singular cardinal $\lambda$. $\square_\lambda$ implies that every stationary subset of $\lambda^+$ admits a non-reflecting stationary subset. $\blacksquare$

We conclude this post by providing proofs to several easy facts (some of which were mentioned in the introduction).

Fact. if $\lambda$ is a regular cardinal, then any subgraph of $EH(\lambda)$ of size $<\lambda$ is countably chromatic.
Proof. Suppose that $\lambda$ is a regular cardinal, and $\mathcal F$ is a family of less than $\lambda$ many functions from $\lambda$ to $\omega$. For $f,g\in\mathcal F$ such that $f\bot g$, denote $$\Delta(f,g):=\min\{ \gamma<\lambda\mid \text{dom}(f\cap g)\subseteq\gamma\}.$$ Let $\gamma:=\sup\{ \Delta(f,g)\mid f,g\in\mathcal F, f\bot g\}$. As $|\mathcal F|<\text{cf}(\lambda)$, we get that $\gamma<\lambda$. Define $c:\mathcal F\rightarrow\omega$ by letting $c(f):=f(\gamma)$ for all $f\in\mathcal F$. Now, if $f,g\in\mathcal F$ and $f\bot g$, then $f(\gamma)\neq g(\gamma)$ for all $\gamma\ge\Delta(f,g)$ and hence $c$ is a chromatic coloring. $\blacksquare$

Fact. If $(G,E)$ is a graph of size $\lambda$ such that every subgraph of size $<\lambda$ is countably chromatic, then the chromatic number of $EH(\lambda)$ is $\ge$ the chromatic number of $(G,E)$.
Proof. Without loss of generality, $G=\lambda$. For every $\gamma<\lambda$, let $c_\gamma:\gamma\rightarrow\omega\setminus\{0\}$ exemplify that $(\gamma,E\cap[\gamma]^2)$ is countably chromatic.
Next, for all $\alpha<\lambda$, define $f_\alpha\in{}^\lambda\omega$ by letting for all $\gamma<\lambda$:$$f_\alpha(\gamma):=\begin{cases}0,&\gamma<\alpha\\g_{\gamma+1}(\alpha),&\text{otherwise}\end{cases}.$$Note that if $\{\alpha,\beta\}\in E$, then $f_\alpha\bot f_\beta$ (indeed, $\Delta(f_\alpha,f_\beta)\le\min\{\alpha,\beta\}$). So, $\alpha\mapsto f_\alpha$ is a graph homomorphism, and the chromatic number of $EH(\lambda)$ is $\ge$ the one of $(G,E)$. $\blacksquare$

Corollary. The chromatic number of $EH(\aleph_1)$ is at least $\aleph_1$. $\blacksquare$

Fact. The chromatic number of $EH(\aleph_0)$ equals $2^{\aleph_0}$.
Proof. Clearly, the chromatic number of $EH(\aleph_0)$ is $\le2^{\aleph_0}$. Next, fix an injection $\psi:[\omega]^{<\omega}\rightarrow\omega$. For every $X\subseteq\omega$, let $f_X:\omega\rightarrow\omega$ be the function defined by $f_X(n):=\psi(X\cap n)$, for all $n<\omega$. Notice that if $X,Y$ are distinct infinite subsets of $\omega$, then $f_X\bot f_Y$. It follows that any chromatic coloring of $EH(\omega)$ must be injective over $\{ f_X\mid X\in[\omega]^\omega\}$, and hence the chromatic number of $EH(\omega)$ equals $2^{\aleph_0}$. $\blacksquare$

Fact. Suppose that $\mathbb T:=\langle T,\le\rangle$ is an $\aleph_1$-tree. Let $G(\mathbb T)$ denote the comparability graph of $\mathbb T$. If $G(\mathbb T)$ is countably chromatic, then $T$ is a special Aronszajn tree.
Proof. Recall that $G(\mathbb T)=(T,E)$, where $E=\{ \{x,y\}\in[T]^2\mid x\le y\text{ or }y\le x\}$. Now, if $C\subseteq T$ is a chain, then any chromatic coloring of $G(\mathbb T)$ must be injective over $C$. In particular, if $G(\mathbb T)$ is countably chromatic, then $\mathbb T$ is Aronszajn. Moreover, if $c:T\rightarrow\omega$ is a chromatic coloring, then it is easy to construct an order-preserving function $o:T\rightarrow\mathbb Q$. Simply define $o$ for all $x\in T$ by induction on $c(x)$, while insuring that $o[c^{-1}\{n\}]$ is finite for all $n<\omega$. $\blacksquare$

Trivia. Todorcevic proved that after Levy collapsing a supercompact to $\omega_2$, the following statement holds in the extension. If $\langle T,\le\rangle$ is a tree for which $G(T,\le)$ is not countably chromatic, then some subtree $T’\in[T]^{\aleph_1}$ already has the property that $G(T’,\le)$ is not countably chromatic.

Waiting for the Polymath revolution — thoughts from a bystander

Peter Krautzberger — Mon, 30 Apr 2012 22:43:18 +0000

Tim Gowers has hinted at a revival of the fifth Polymath project. Which brings something back from the bottom of my draft folder.

Let’s talk about Polymath

If you haven’t heard of the Polymath project, then, hm, well… anyway, here’s …

The countable models of ZFC, up to isomorphism, are linearly ordered by the submodel relation; indeed, every countable model of ZFC, including every transitive model, is isomorphic to a submodel of its own $L$, New York, 2012

Joel David Hamkins — Mon, 30 Apr 2012 14:00:23 +0000

This will be a talk on May 18, 2012 for the CUNY Logic Workshop on some extremely new work. The proof uses finitary digraph combinatorics, including the countable random digraph and higher analogues involving uncountable Fraisse limits, the surreal numbers and the hypnagogic digraph.

The story begins with Ressayre’s remarkable 1983 result that if $M$ is any nonstandard model of PA, with $\langle\text{HF}^M,{\in^M}\rangle$ the corresponding nonstandard hereditary finite sets of $M$, then for any consistent computably axiomatized theory $T$ in the language of set theory, with $T\supset ZF$, there is a submodel $N\subset\langle\text{HF}^M,{\in^M}\rangle$ such that $N\models T$. In particular, one may find models of ZFC or even ZFC + large cardinals as submodels of $\text{HF}^M$, a land where everything is thought to be finite. Incredible! Ressayre’s proof uses partial saturation and resplendency to prove that one can find the submodel of the desired theory $T$.

My new theorem strengthens Ressayre’s theorem, while simplifying the proof, by removing the theory $T$. We need not assume $T$ is computable, and we don’t just get one model of $T$, but rather all models—the fact is that the nonstandard models of set theory are universal for all countable acyclic binary relations. So every model of set theory is a submodel of $\langle\text{HF}^M,{\in^M}\rangle$.

Theorem.(JDH) Every countable model of set theory is isomorphic to a submodel of any nonstandard model of finite set theory. Indeed, every nonstandard model of finite set theory is universal for all countable acyclic binary relations.

The proof involves the construction of what I call the countable random $\mathbb{Q}$-graded digraph, a countable homogeneous acyclic digraph that is universal for all countable acyclic digraphs, and proving that it is realized as a submodel of the nonstandard model $\langle M,\in^M\rangle$. Having then realized a universal object as a submodel, it follows that every countable structure with an acyclic binary relation, including every countable model of ZFC, is realized as a submodel of $M$.

The proof, in brief: for every countable acyclic digraph, consider the partial order induced by the edge relation, and extend this order to a total order, which may be embedded in the rational order $\mathbb{Q}$. Thus, every countable acyclic digraph admits a $\mathbb{Q}$-grading, an assignmment of rational numbers to nodes such that all edges point upwards. Next, one can build a countable homogeneous, universal, existentially closed $\mathbb{Q}$-graded digraph, simply by starting with nothing, and then adding finitely many nodes at each stage, so as to realize the finite pattern property. The result is a computable presentation of what I call the countable random $\mathbb{Q}$-graded digraph $\Gamma$. If $M$ is any nonstandard model of finite set theory, then we may run this computable construction inside $M$ for a nonstandard number of steps. The standard part of this nonstandard finite graph includes a copy of $\Gamma$. Furthermore, since $M$ thinks it is finite and acyclic, it can perform a modified Mostowski collapse to realize the graph in the hereditary finite sets of $M$. By looking at the sets corresponding to the nodes in the copy of $\Gamma$, we find a submodel of $M$ that is isomorphic to $\Gamma$, which is universal for all countable acyclic binary relations. So every model of ZFC isomorphic to a submodel of $M$.

The proof idea adapts, with complications, to the case of well-founded models, via the countable random $\lambda$-graded digraph, as well as the internal construction of what I call the hypnagogic digraph, a proper class homogeneous surreal-numbers-graded digraph, which is universal for all class acyclic digraphs.

Theorem.(JDH) Every countable model $\langle M,\in^M\rangle$ of ZFC, including the transitive models, is isomorphic to a submodel of its own constructible universe $\langle L^M,\in^M\rangle$. In other words, there is an embedding $j:M\to L^M$ that is quantifier-free-elementary.

The proof is guided by the idea of finding a universal submodel inside $L^M$. The embedding $j$ is constructed completely externally to $M$.

Corollary.(JDH) The countable models of ZFC are linearly ordered and even well-ordered, up to isomorphism, by the submodel relation. Namely, any two countable models of ZFC with the same well-founded height are bi-embeddable as submodels of each other, and all models embed into any nonstandard model.

The work opens up numerous questions on the extent to which we may expect in ZFC that $V$ might be isomorphic to a subclass of $L$. To what extent can we expect to have or to refute embeddings $j:V\to L$, elementary for quantifier-free assertions?

I am preparing the article now, which I hope will be ready soon.

The omega one of infinite chess, New York, 2012

Joel David Hamkins — Sat, 28 Apr 2012 23:10:56 +0000

This will be a talk on May 18, 2012 for the CUNY Set Theory Seminar.

Infinite chess is chess played on an infinite edgeless chessboard. The familiar chess pieces move about according to their usual chess rules, and each player strives to place the opposing king into checkmate. The mate-in-$n$ problem of infinite chess is the problem of determining whether a designated player can force a win from a given finite position in at most $n$ moves. A naive formulation of this problem leads to assertions of high arithmetic complexity with $2n$ alternating quantifiers—there is a move for white, such that for every black reply, there is a countermove for white, and so on. In such a formulation, the problem does not appear to be decidable; and one cannot expect to search an infinitely branching game tree even to finite depth. Nevertheless, in joint work with Dan Brumleve and Philipp Schlicht, confirming a conjecture of myself and C. D. A. Evans, we establish that the mate-in-$n$ problem of infinite chess is computably decidable, uniformly in the position and in $n$. Furthermore, there is a computable strategy for optimal play from such mate-in-$n$ positions. The proof proceeds by showing that the mate-in-$n$ problem is expressible in what we call the first-order structure of chess, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable. An equivalent account of the result arises from the realization that the structure of chess is interpretable in the standard model of Presburger arithmetic $\langle\mathbb{N},+\rangle$. Unfortunately, this resolution of the mate-in-$n$ problem does not appear to settle the decidability of the more general winning-position problem, the problem of determining whether a designated player has a winning strategy from a given position, since a position may admit a winning strategy without any bound on the number of moves required. This issue is connected with transfinite game values in infinite chess, and the exact value of the omega one of chess $\omega_1^{\rm chess}$ is not known. I will also discuss recent joint work with C. D. A. Evans and W. Hugh Woodin showing that the omega one of infinite positions in three-dimensional infinite chess is true $\omega_1$: every countable ordinal is realized as the game value of such a position.

article | slides

Shelah’s approachability ideal (part 1)

Assaf Rinot — Wed, 25 Apr 2012 03:13:38 +0000

Given an infinite cardinal $\lambda$, Shelah defines an ideal $I[\lambda]$ as follows.

Definition (Shelah, implicit in here). A set $S$ is in $I[\lambda]$ iff $S\subseteq\lambda$ and there exists a collection $\{ \mathcal D_\alpha\mid\alpha<\lambda\}\subseteq\mathcal [\mathcal P(\lambda)]^{<\lambda}$, and some club $E\subseteq\lambda$, so that for every $\delta\in S\cap E$, there exists a cofinal subset $A_\delta\subseteq\delta$ such that:

$\text{otp}(A_\delta)<\delta$ (in particular, $\delta$ is singular);
for every $\gamma<\delta$, there exists some $\alpha<\delta$ such that $A_\delta\cap\gamma\in\mathcal D_\alpha$.

In other words, $\bigcup_{\alpha<\delta}\mathcal D_\alpha$ contains all the initial segments of some small cofinal subset, $A_\delta$, of $\delta$.

Easy observations. 1) $E^\lambda_\omega\in I[\lambda]$;
2) if $\mu^{<\kappa}<\lambda$ for all $\mu<\lambda$, then $E^\lambda_\kappa\in I[\lambda]$;
3) if $S\subseteq\lambda$ is nonstationary, then $S\in I[\lambda]$;
4) if $\square_\lambda$ holds, then $I[\lambda^+]=\mathcal P(\lambda^+)$. (the definition of $\square_\lambda$ may be found in here.)
Proof hints. 1) Let $\mathcal D_\alpha:=[\alpha]^{<\omega}$ for all $\alpha<\lambda$.
2) Let $\mathcal D_\alpha:=[\alpha]^{<\kappa}$ for all $\alpha<\lambda$.
3) Take a club $E$ which is disjoint from $S$.
4) If $\langle C_\alpha\mid\alpha<\lambda^+\rangle$ is a $\square_\lambda$-sequence, then simply let $\mathcal D_\alpha:=\{\text{acc}(C_\alpha)\}\cup[C_\alpha]^{<\omega}$ for all $\alpha<\lambda^+$, and restrict your attention to $E:=\text{acc}(\lambda^+)$. $\square$

So, $I[\lambda^+]$ is rather trivial in the presence of $\square_\lambda$. On the other hand, it is a very sophisticated result of Mitchell, that $I[\lambda^+]$ may behave as the other extreme:

Theorem (Mitchell, 2009). Starting with a cardinal $\kappa$ which is $\kappa^+$-Mahlo, in some forcing extension, $I[\aleph_2]$ has the property that $S\cap E^{\aleph_2}_{\aleph_1}$ is nonstationary for every $S\in I[\aleph_2]$. $\square$

In other words, $I[\kappa^{++}]$ restricted to cofinality $\kappa^+$ may coincide with nonstationary ideal $NS^{\kappa^{++}}_{\kappa^+}$. Next, let us ask what about $I[\kappa^{++}]$ restricted to cofinality $\kappa$? it turns out that we are back on track here, and this follows from the trivial fact that $\kappa^{++}$ is a successor of a regular cardinal, together with the following less-trivial proposition:

Proposition (Shelah, 1991). $E^{\lambda^+}_{<\lambda}\in I[\lambda^+]$ for every regular cardinal $\lambda$.
Proof. Of course, we may assume that $\lambda>\aleph_0$. For every ordinal $\alpha<\lambda^+$, fix an injection $d_\alpha:\alpha\rightarrow\lambda$. Notice that for every $\alpha<\delta<\lambda^+$:

$\forall i<\lambda\exists j\in[i,\lambda)$ such that $d_\delta^{-1}[i]\cap\alpha\subseteq d_\alpha^{-1}[j]$;
$\forall i<\lambda\exists j\in[i,\lambda)$ such that $d_\delta^{-1}[j]\cap\alpha\supseteq d_\alpha^{-1}[i]$.

It follows that for all $\alpha<\delta<\lambda^+$, the following set is a club in $\lambda$: $$C_\alpha^\delta:=\{ i<\lambda\mid d_\delta^{-1}[i]\cap\alpha=d_\alpha^{-1}[i]\}.$$

Next, for all $\alpha<\lambda^+$, let $$\mathcal D_\alpha:=\{ d_\alpha^{-1}[i]\cap\gamma\mid \gamma\le\alpha,i<\lambda\}.$$ Clearly, $\{ \mathcal D_\alpha\mid \alpha<\lambda^+\}\subseteq[\mathcal P(\lambda^+)]^{\le\lambda}$. We now fix an arbitrary limit ordinal $\delta\in E^{\lambda^+}_{<\lambda}$, and show that $\bigcup_{\alpha<\delta}\mathcal D_\alpha$ contains all the initial segments of some small cofinal subset of $\delta$.
Let $u$ be a cofinal subset of $\delta$ of minimal order-type. In particular, $|u|<\lambda$, and hence $C:=\bigcap_{\alpha\in u}C^\delta_\alpha$ is a club in $\lambda$. Put $i:=\min(C\setminus\sup(d_\delta[u])+1)$, and $A_\delta:=d^{-1}_\delta[i]$. Then:

$A_\delta\subseteq\delta$;
$u\subseteq A_\delta$, and hence $\sup(A_\delta)=\delta$;
$|A_\delta|=|i|$, and hence $|A_\delta|<\lambda$;
for every $\alpha\in u$, as $i\in C^\delta_\alpha$, we have $ A_\delta\cap\alpha=d_\delta^{-1}[i]\cap\alpha=d_\alpha^{-1}[i]$;
for every $\gamma<\delta$, letting $\alpha:=\min(u\setminus\gamma)$, we get that $$A_\delta\cap\gamma=A_\delta\cap\alpha\cap\gamma=d_\alpha^{-1}[i]\cap\gamma\in \mathcal D_\alpha.$$

This completes the proof. $\square$

We now arrive to the main result of today’s post:

Theorem (Shelah, 1993). If $\kappa$ is a regular cardinal, and $\kappa^+<\lambda$, then $I[\lambda]$ contains a stationary subset of $E^\lambda_\kappa$.
Proof. We already know that $E^\lambda_\omega\in I[\lambda]$. We also know that $E^\lambda_\kappa\in I[\lambda]$ in the case $\lambda=\kappa^{++}$. Thus, we shall assume that $\aleph_0<\kappa<\kappa^{++}<\lambda$.
In an earlier post, we proved that $E^{\aleph_3}_{\aleph_1}$ carries a club guessing sequence, and since here $\kappa$ is regular and uncountable, the same argument shows that $E^{\kappa^{++}}_\kappa$ carries a club-guessing sequence. Thus, let us fix such a club-guessing sequence $\overrightarrow C=\langle C_\beta\mid \beta\in E^{\kappa^{++}}_\kappa\rangle$. Next, fix a large enough regular cardinal $\theta$, and let $\overrightarrow M=\langle M_\alpha\mid \alpha<\lambda\rangle$ be an increasing sequence of elementary submodels of $(\mathcal H_\theta,\in)$ such that for all $\alpha<\lambda$:

$|M_\alpha|<\lambda$;
$\kappa^{++}+\alpha+1\subseteq M_{\alpha+1}$;
$\overrightarrow C,\lambda\in M_\alpha$.

We shall now make a somewhat naive move, and simply let $\mathcal D_\alpha:=M_\alpha\cap\mathcal P(\lambda)$ for all $\alpha<\lambda$. Since $\mathcal D_\alpha\in[\mathcal P(\lambda)]^{<\lambda}$ for all $\alpha<\lambda$, the following set $S$ is obviously in $I[\lambda]$,$$S:=\{ \delta\in E^\lambda_\kappa\mid \exists A_\delta\subseteq\delta(\text{otp}(A_\delta)<\delta=\sup(A_\delta),\forall\gamma<\delta\exists\alpha<\delta[A_\delta\cap\gamma\in\mathcal D_\alpha])\}.$$

On the other hand, the next statement is not entirely obvious.

Subclaim. $S$ is stationary.
Proof. Given a club $D\subseteq\lambda$, we shall seek some $\delta\in S\cap D$. Wlog, $\min(D)>\kappa$.
Let $\overrightarrow N:=\langle N_i\mid i<\kappa^{++}\rangle$ be a sequence of of elementary submodels of $(\mathcal H_\theta,\in)$ such that for all $i<\kappa^{++}$:

$|N_i|=\kappa^{++}$, with $\kappa^{++}\subseteq N_{i}$;
$\langle N_j\mid j\le i\rangle\in N_{i+1}$;
$\overrightarrow M, \overrightarrow C,D,\lambda\in N_i$;
$N_i=\bigcup_{j<i}N_j$ whenever $i$ is a limit ordinal.

For all $i<\kappa^{++}$, as $D,\lambda\in N_i$ and $\sup(D)=\lambda$, we get by elementarity that $\sup(N_i\cap D)=\sup(N_i\cap\lambda)$. Recalling that $D$ is closed, we infer that $\sup(N_i\cap\lambda)\in D$ for all $i<\kappa^{++}$. Thus, it make sense to define a function $f:\kappa^{++}\rightarrow D$ by letting: $$f(i):=\sup(N_i\cap\lambda),\quad(i<\kappa^{++}.)$$
By the second and fourth defining properties of $\overrightarrow N$, we get that $f$ is increasing and continuous, and that $f\restriction\tau\in N_{\tau+1}$ for all $\tau<\kappa^{++}$. Put $\epsilon:=\sup(f[\kappa^{++}])$. Since $\kappa^{++}+\epsilon+1\subset M_{\epsilon+1}$, we get (by elementarity) the existence of some strictly-increasing and cofinal function $g:\kappa^{++}\rightarrow\epsilon$ that lies in $M_{\epsilon+1}$.
Since $f$ and $g$ are both continuous and cofinal in $\epsilon$, a standard back-and-fourth argument yields the existence of a club $c\subseteq\kappa^{++}$ for which $f\restriction c=g\restriction c$. Since $\overrightarrow C$ is a club guessing sequence, we may find some $\beta\in E^{\kappa^{++}}_\kappa$ such that $C_\beta\subseteq c$. Put $\delta:=f(\beta)$ and $A_\delta:=f[c_\beta]$. Then $\delta\in D\cap E^\lambda_\kappa$, and $A_\delta$ is a cofinal subset of $\delta$ of order-type $\kappa<\min(D)\le\delta$. Thus, we are left with showing that $A_\delta\cap\gamma\in\bigcup_{\alpha<\delta}\mathcal D_\alpha$ for all $\gamma\in A_\delta$.
Fix $\gamma\in A_\delta$, and let $\tau:=f^{-1}(\gamma)$. Then:

$\tau+1<\beta$;
$A_\delta\cap\gamma=g[c_\beta\cap\tau]$, and the latter belongs to $M_{\epsilon+1}$, since $g,\overrightarrow C,\beta,\tau\in M_{\epsilon+1}$ ;
$A_\delta\cap\gamma=(f\restriction\tau)[c_\beta]$, and the latter belongs to $N_{\tau+1}$ since $f\restriction\tau,\overrightarrow C,\beta\in N_{\tau+1}$.

Denote $A:=A_\delta\cap\gamma$. Then $A\in M_{\epsilon+1}$, and $A,\overrightarrow M\in N_{\tau+1}$. Consequently: $$N_{\tau+1}\models \exists\alpha<\lambda(A\in M_\alpha).$$ It follows that there exists some $\alpha<\sup(N_{\tau+1}\cap\lambda)=f(\tau+1)<f(\beta)=\delta$ such that $A\in M_\alpha$. In particular, $A_\delta\cap\gamma\in\bigcup_{\alpha<\delta}\mathcal D_\alpha$. $\square$

So, $S$ is an example of a stationary subset of $E^\lambda_\kappa$ that belongs to $I[\lambda]$. $\square$

1. Notice that in the above proof, we could have replaced $\kappa^{++}$ with any regular cardinal $\mu<\lambda$ for which $E^\mu_\kappa$ carries a club-guessing sequence. In particular, the above proof shows:

Theorem (Shelah). If $\kappa<\kappa^+<\mu<\lambda$ are all regular cardinals, then there exists a subset $S\subseteq E^\lambda_\kappa$ such that:

$S\in I[\lambda]$, and $S$ reflects in the following sense:
$\{\delta\in E^\lambda_\mu\mid S\cap\delta\text{ is stationary}\}$ is stationary in $\lambda$. $\square$

2. The proof may be adapted to show that if $\kappa<\kappa^+<\lambda$ are regular, then there exists a sequence $\langle C_\delta\mid \delta\in E^\lambda_\kappa\rangle$, and an enumeration $\{ \mathcal D_\alpha\mid\alpha<\lambda\}\subseteq[\mathcal P(\lambda)]^{<\lambda}$, such that for every club $D\subseteq\lambda$, there exists $\delta\in\text{acc}(D)$ with:

$\text{otp}(C_\delta)=\text{cf}(\delta)=\kappa$;
$C_\delta$ is a club subset of of $D\cap\delta$;
for every $\gamma<\delta$, there exists $\alpha<\delta$ such that $C_\delta\cap\gamma\in\mathcal D_\alpha$.

Automorphisms of corona algebras, and group cohomology

Samuel Coskey — Tue, 24 Apr 2012 15:27:35 +0000

With Ilijas Farah. (arXiv)

This article gives a generalization of a (relatively) recent result concerning the Calkin algebra. It is just one of a number of recent work that mixes set theory and the study of non-separable C*-algebras. I’m really grateful to Ilijas for pulling me into this project, which belongs essentially entirely to him. Thanks to his enthusiasm, I have much new understand and respect for this remarkable subject.

First, let me define the Calkin algebra. Let $H$ be a separable Hilbert space, and $\mathcal K(H)$ denote the algebra of compact operators on $H$: the norm-closure of the algebra of operators with finite-dimensional range. Then $\mathcal K(H)$ is a two-sided ideal in the algebra $B(H)$ of all bounded linear operators on $H$, and the Calkin algebra is just the quotient $B(H)/\mathcal K(H)$.

In 2007, Phillips and Weaver proved that CH implies that the Calkin algebra has an outer automorphism (i.e., one that isn’t just conjugation by a single element). In this paper, we establish the same result for a much more general class of quotient structures called corona algebras.

Here, if $A\subset B(H)$ then the multiplier algebra of $A$ is the set $M(A)=\set{m\in B(H):mA\subset A\text{ and }Am\subset A}$. The corona of $A$ is the quotient $M(A)/A$. It is easy to see that the corona of $\mathcal K$ is exactly the Calkin algebra, and that the corona of $C(X)$ is exactly $C(\beta X\smallsetminus X)$, the algebra of continuous functions on the corona of $X$. This is where the corona derives its name. We prove:

Theorem. Assume CH holds. If $A$ is separable and is either simple or stable, then the corona of $A$ has $2^{2^{\aleph_0}}$ many automorphisms.

This hypotheses on $A$ are not the weakest possible, but some hypothesis is certainly necessary. There are several algebras $A$ for which it is known that all automorphisms are inner, and there are several cases left open.

Our proof is based on Farah’s proof of the result of Phillips and Weaver in the case of the Calkin Algebra. Loosely speaking, we stratify the corona into $\aleph_1$ many layers and then build a complete binary tree of height $\aleph_1$ consisting of partial automorphisms. Here the $\alpha\th$ level of the tree corresponds to the $\alpha\th$ stage in the stratification. We take care to ensure that compatible elements of the tree correspond to compatible partial automorphisms, and, by small perturbations at each step, that incompatible elements correspond to distinct partial automorphisms. It follows that the set of branches through the tree give $2^{\aleph_1}$ distinct automorphisms, and assuming CH, this is the desired number.

In the process of the proof, we uncovered a small connection with category theory. I will not elaborate on it here, but rather just note that our proof can be viewed as a computation of the cardinality of a particular derived inverse limit.

Lastly, I should mention that we do not address the (much more important) complementary question. Namely, Farah has also shown that Todorčević’s Axiom (also called OCA) implies that the Calkin algebra has no outer automorphisms. It would be very interesting to generalize this much harder work to the case of corona algebras as well.

yay, I’m an editor at ScienceSeeker.org

Peter Krautzberger — Sat, 21 Apr 2012 19:07:47 +0000

Apparently, I totally forgot to announce this anywhere on my interwebz!

With its major code update three weeks ago, scienceseeker.org introduced an editorial system where editors can mark particular posts and leave a small note — and I was invited …

Review: Is classical set theory compatible with quantum experiments?

Assaf Rinot — Thu, 19 Apr 2012 02:23:00 +0000

Yesterday, I attended a talk at the Quantum Foundations seminar at the beautiful Perimeter Institute for Theoretical Physics (Waterloo, Ontario).
The (somewhat provocative) title of the talk was “Is Classical Set Theory Compatible with Quantum Experiments?”, and the speaker was Radu Ionicioiu. Here are the links to the slides and videotape.

To make a long story short, the speaker addresses the Sakurai’s Bell inequality that can be thought of yielding a decomposition of a certain set $S(a_x,b_y)$ into two parts $S(a_x,b_y,c_+)$ and $S(a_x,b_y,c_{-})$, each of which being unknown/non-understood/non-definite, while each of these sets can be described as the set of all elements of the original set obeying a certain property. Thus:

“Recalling that ZFC‘s axiom of separation asserts that if $S$ is a set, and $\mathcal P$ is a property, then $\{ x\in S\mid \mathcal P(x)\text{ holds}\}$ is a set, we conclude that the results of the Sakurai-Bell experiment are incompatible with classical set theory (or incompatible with even more foundational logical rules, such as the law of excluded middle that implies that the union of the two sets would resurrect $S(a_x,b_y)$).”

Now, here are briefly my thoughts on this argument:

The axiom of separation does not apply to arbitrary properties $\mathcal P(x)$. It only applies to formulas $\mathcal P(x)$ in the language of set theory! (Recall the paradox of the heap.)
The discussed set (that admits an unclear decomposition) is finite – one of the participants of the seminar mentioned that it contains around 20k many elements. Partitioning a 20k-sized-set got nothing to do with ZFC!
The superadditivity idea on slide 43 is a consequence of a confusion between undefined sets and the empty set. (some sets are undefined, some undefined objects are simply not sets, and in any case, undefined object is not necessarily the empty set).
On slide 48, the speaker suggests that just as that the elimination of the fifth postulate from Euclidean geometry give rise to fertile (“non-Euclidean”) theories, the elimination of the separation scheme from ZFC could give rise to valuable theories. While this may sounds plausible, one should take into account that the fifth postulate is a restrictive axiom (hence, removing it, gives more freedom), while the separation axiom is responsible for the “existence” of sets (hence, removing it, kills many arguably-reasonable sets). To be even more picky, the axiom of separation anyway follows from the axiom of replacement, provided that an empty set exists.
Nine years ago, I attended a talk by Guy Gildor, on his master thesis concerning a development of Set Theory within Fuzzy Logic. His work is perfectly rigorous, and may be found more convenient as foundations for quantum physics. I wonder if anyone ever considered that..?

Models of $\rm{ZFC}^-$ that are not definable in their set forcing extensions

Victoria Gitman — Wed, 18 Apr 2012 21:42:38 +0000

$
\newcommand{\p}{\mathbb{P}}
\newcommand{\ZFC}{\rm{ZFC}}
\newcommand{\her}[1]{H_{{#1}^+}}
$

This is a talk at the CUNY Set Theory Seminar, May 4, 2012.

It took four decades since the invention of forcing for set theorists and to ask (and answer) what post factum seems as one of the most natural questions regarding forcing. Is the ground model a definable class of its set forcing extensions? Using techniques developed by Hamkins, Laver provided a positive answer in [1]. The result is also due independently to Woodin from about the same time [2].

(Laver, Woodin, Hamkins)

Suppose $V$ is a model of $\ZFC$, $\p\in V$ is a forcing notion, and $G\subseteq\p$ is $V$-generic. Then in $V[G]$, the ground model $V$ is uniformly definable from the parameter $\mathcal P(|\p|)^V$.

Not only is the ground model always a definable class of its set forcing extensions, but the uniform definition uses a ground model parameter. This is a good time to point out that the requirement that $\p$ is a set forcing is a necessary one, since the ground model need not be definable in a class forcing extension. Here is a counterexample hint. Let $\p$ be the class length Easton support product adding a Cohen subset to every regular cardinal. Note that $\p$ is isomorphic to $\p\times \p$ since adding two Cohen reals is the same as adding one. For the full argument, due to Sy Friedman, see this MathOverflow question.

Since, in many contexts, set theorists force over models of $\ZFC^{-}$, the theory $\ZFC$ without the powerset axiom, it is also natural to wonder whether in this case the ground model is definable in its set forcing extensions.

Strange things have been known to happen before once the powerset axiom is removed. It has been a project of Zarach for many years now to convince the readers of his papers that $\ZFC^{-}$ behaves unexpectedly and unintuitively. Zarach showed that without the powerset axiom, the replacement and collection schemes are not equivalent [3], and Hamkins, Johnstone and myself showed that collection is necessary for many of the constructions that set theorists take for granted, particularly in the context of large cardinals, as for example, the Łós ultrapower construction [4]. Thus, we must to be careful to specify that $\ZFC^{-}$ includes collection in place of the replacement scheme. Another unintuitive consequence, pointed out by Zarach (proved by his colleague Szczpaniak), is that without the powerset axiom, the axiom of choice does not imply the well-ordering principle [5]. And so Thomas Johnstone advocates that when defining $\ZFC^-$, we adopt the convention that ${\rm C}$ stands for the well-ordering principle. Most natural examples of $\ZFC^{-}$ models arise as elementary substructures of $\her{\kappa}$, the set of all sets of hereditary size $\leq\kappa$, for some cardinal $\kappa$. These are clearly models of both the collection scheme and the well-ordering principle.

It turns out that a $\ZFC^{-}$ ground model need not be definable in a set forcing extension. Indeed, using different preparatory forcing, one can create numerous such counterexamples of the form $\her{\kappa}$. The downside of the counterexample models is that the powerset of the forcing notion in whose extension they are not definable is a proper class in the model, making the situation much too similar to class forcing to be truly satisfying. These models do have the interesting feature, as observed by Joel Hamkins, of also serving as counterexamples to the standard $\ZFC$ fact that the ground model can never be elementary in its set forcing extension. This should not be overly surprising since the non-elementarity in the $\ZFC$ case is established using the powerset of the forcing notion as a parameter.

So, is there a $\ZFC^{-}$ model that is not definable in an extension by a forcing notion whose powerset is an element of the model? I asked this question to David Asperó recently and he produced a counterexample, using, of all things, an $I_0$-cardinal, a large cardinal that sits pretty much atop the large cardinal hierarchy. An $I_0$-embedding is an elementary embedding $j:L(V_{\lambda+1})\to L(V_{\lambda+1})$ with critical point $\kappa<\lambda$ and $\kappa$ is an $I_0$-cardinal. The existence of $I_0$-cardinals pushes right up against the Kunen Inconsistency, the existence of an elementary embedding $j:V\to V$. Incidentally, anyone interested to learn about $I_0$-cardinals should read Vincenzo Dimonte’s excellent notes. Do we really need large cardinals, especially ones this large, to produce the such a counterexample? One should hope not, but absolutely nothing is known about it!

For details, take a look at the lecture notes.

[1] R. Laver, “Certain very large cardinals are not created in small forcing
extensions,” Ann. Pure Appl. Logic, vol. 149, iss. 1-3, pp. 1-6, 2007.
[Bibtex]

@article {laver:groundmodel,
AUTHOR = {Laver, Richard},
TITLE = {Certain very large cardinals are not created in small forcing
extensions},
JOURNAL = {Ann. Pure Appl. Logic},
FJOURNAL = {Annals of Pure and Applied Logic},
VOLUME = {149},
YEAR = {2007},
NUMBER = {1-3},
PAGES = {1--6},
ISSN = {0168-0072},
CODEN = {APALD7},
MRCLASS = {03E55 (03E35)},
MRNUMBER = {2364192 (2009e:03099)},
MRREVIEWER = {Paul Bradley Larson},
DOI = {10.1016/j.apal.2007.07.002},
URL = {http://dx.doi.org/10.1016/j.apal.2007.07.002},
}

[2] H. Woodin, “Recent developments on Cantor’s Continuum Hypothesis,” in Proceedings of the continuum in Philosophy and Mathematics, 2004.
[Bibtex]

@inproceedings {woodin:groundmodel,
AUTHOR = {Woodin, Hugh},
TITLE = {Recent developments on {C}antor's {C}ontinuum {H}ypothesis},
BOOKTITLE = {Proceedings of the continuum in {P}hilosophy and {M}athematics},
PUBLISHER = {Carlsberg Academy, {C}oppenhagen},
YEAR = {2004},
}

[3] A. M. Zarach, “Replacement $\nrightarrow$ collection.” Berlin: Springer, 1996, vol. 6, pp. 307-322.
[Bibtex]

@incollection {Zarach1996:ReplacmentDoesNotImplyCollection,
AUTHOR = {Zarach, Andrzej M.},
TITLE = {Replacement {$\nrightarrow$} collection},
BOOKTITLE = {G\"odel '96 ({B}rno, 1996)},
SERIES = {Lecture Notes Logic},
VOLUME = {6},
PAGES = {307--322},
PUBLISHER = {Springer},
ADDRESS = {Berlin},
YEAR = {1996},
MRCLASS = {03E30 (03E35)},
MRNUMBER = {1441120 (98g:03120)},
}

[4] V. Gitman, J. D. Hamkins, and T. A. Johnstone, “What is the theory ZFC without power set?.”
[Bibtex]

@ARTICLE{zfcminus:gitmanhamkinsjohnstone,
AUTHOR= {Victoria Gitman and Joel David Hamkins and Thomas A. Johnstone},
TITLE= {What is the theory {ZFC} without power set?},
NOTE= {Submitted},
PDF={http://boolesrings.org/victoriagitman/files/2011/10/ZFC-.pdf},
EPRINT={1110.2430}}

[5] A. Zarach, “Unions of ${\rm ZF}^{-}$-models which are themselves
${\rm ZF}^{-}$-models.” Amsterdam: North-Holland, 1982, vol. 108, pp. 315-342.
[Bibtex]

@incollection {zarach:unions_of_zfminus_models,
AUTHOR = {Zarach, Andrzej},
TITLE = {Unions of {${\rm ZF}^{-}$}-models which are themselves
{${\rm ZF}^{-}$}-models},
BOOKTITLE = {Logic {C}olloquium '80 ({P}rague, 1980)},
SERIES = {Stud. Logic Foundations Math.},
VOLUME = {108},
PAGES = {315--342},
PUBLISHER = {North-Holland},
ADDRESS = {Amsterdam},
YEAR = {1982},
MRCLASS = {03C62 (03E35)},
MRNUMBER = {673801 (84h:03086)},
MRREVIEWER = {M. Dubiel},
}

Math hangout: like water for calculators

Samuel Coskey — Wed, 18 Apr 2012 02:37:33 +0000

Calculators perform computations by purposefully controlling the flow of electricity. On monday we tried to understand this just a little by replacing electricity with something a little more concrete: water.

Specifically, we asked how you build an adder—a machine which adds two whole numbers. This means we have to decide how to represent numbers using water. We decided that a number $n$ would be represented by $n$ pipes with flowing water. So a machine that can add numbers between 0 and 2 would look like this:

The diagram shows “one plus two equals three”. There are other ways to pose the same problem. Avital noticed that the machine is essentially just a counter, hat is, it counts how many of its input pipes are flowing. Another observation is that it is just a sorter, that is, it sorts all of the water to the left-most pipes.

Our group came up with several solutions.

Gravity. My picture doesn’t do it justice (next time I’ll screen capture our work), but we came up with a pretty convincing arrangement along these lines.

The idea is that the water would fill the lower pipes first, thereby arranging themselves from left to right as we wanted. Once we were satisfied by this, we added a new condition: the solution should work horizontally, so that gravity would not play a role.

Pressure. The next suggestion was to put a pressure-sensitive gate on each output pipe. This gate would allow water to flow through it only when the pressure inside the machine reaches a certain point. The calibration of the gates would vary.

In this picture, we are assuming that water is flowing through each input pipe at 1 unit of pressure. If there is any water coming in, then we want it to flow out of the left-most pipe, so there is no need for a pressure gate there. If two input pipes are flowing, then one pipe’s worth will flow out the left-most pipe leaving behind a positive amount of pressure in the box. The gate at the second pipe will open before any other gate, and when it does it will relieve this pressure and no more gates will open. We used the same reasoning to show it works when water flows through three or four input pipes.

Logic. We didn’t have time to discuss it further, but at the end I admitted what I had been thinking. If you build the simplest adder, the one which can add numbers up to only 1, then you have already computed two logical operations:

(Here v means “or” and ^ means “and”.) It is possible to combine these simple operations again and again to build bigger operations. The advantage of re-using simple operations is that you use a small number of distinct components many times. This is as opposed to the pressure solution, where we needed as many distinct components as we had pipes.

For instance, to go to three pipes you can write:

How can you combine simple components to obtain these three logical operations? And how can you generalize this idea to even more pipes?

state of foundational research

Peter Krautzberger — Sat, 14 Apr 2012 02:21:01 +0000

I was catching up on some fun google-reader reading today but found with a depressing combination of content.

To be

The title of this post is taken from this math.SE question which boils down to:

Is it true? Is it
…

How Nwd captures Meager

Samuel Coskey — Fri, 13 Apr 2012 23:15:29 +0000

Let Nwd denote the collection of closed, nowhere dense subsets of $2^{\mathbb N}$, and Meager the $\sigma$-ideal generated by Nwd. Then all of the cardinal invariants of Meager can be expressed in terms of the inclusion partial ordering on Nwd.

To begin, every partial order P carries two very important invariants:

cf(P) = the least cardinality of a cofinal subset of P
add(P) = the least cardinality of an unbounded subset of P

These invariants are preserved under an appropriate notion of morphism. Namely, P is said to be Tukey reducible to Q if there is a function $f\colon P\to Q$ such that $f^{-1}$ carries bounded sets to bounded sets.

Theorem. If P and Q are Tukey bireducible, then cf(P)=cf(Q) and add(P)=add(Q).

(The proof is not difficult.)

Now, I had always thought that Nwd and Meager were so closely related that they should have the same invariants, and in fact, that they would be Tukey bireducible. This intuition is essentially correct, but the real story is just a little bit more technical than that. In the remainder of this post, I will explain this in detail.

My intuition turns out to be half-correct: in fact cf(Nwd)=cf(Meager), but this is not the case for add(). In fact since Meager is a $\sigma$-ideal, add(Meager) is always uncountable, whereas add(Nwd) is countable! (Just think of an enumeration of the rationals….) We can take care of this using an ad-hoc crutch, namely, let:

$\sigma$-add(P) = the least cardinality of a subset of P which cannot be dominated by a countable subset of P

Then it is clear that $\sigma$-add(Nwd) is exactly add(Meager). What is more, $\sigma$-add() is again preserved under Tukey bireducibility and even the weaker (and more suitable) notion of $\sigma$-Tukey reducibility: P is said to be $\sigma$-Tukey reducible to Q if there is a function $f\colon P\to Q$ such that $f^{-1}$ carries bounded sets to sets which are dominated by a countable family.

Theorem. If P and Q are $\sigma$-Tukey bireducible, then cf(P)=cf(Q) (if they are infinite) and $\sigma$-add(P) = $\sigma$-add(Q).

Again, the proof is very direct. This tells us that $\sigma$-Tukey reducibility is just what we need to properly state relationship between Nwd and Meager.

The Main Theorem. Nwd is $\omega$-Tukey equivalent to Meager.

This is the sense in which Nwd captures all of the cardinal invariants of Meager. In the remainder of the post, let me outline a proof of the theorem.

First let’s find a $\sigma$-Tukey map from Meager to Nwd; in fact in this direction we will find a Tukey map. To proceed, we’ll need some notation. For any sequence $\tau\in2^{<\NN}$ and any subset $F\subset 2^\NN$, let $\tau\concat F$ denote the set $\set{\tau\concat x:x\in F}$ and let $F^{-\tau}$ denote the set $\set{x:\tau\concat x\in F}$.

We can now define the map. Begin by fixing a sequence $\tau_n$ of pairwise incomparable elements of $2^{<\NN}$. Then given a meager set $M$, we first write $M\subset\bigcup A_n$ where $A_n$ are in Nwd. Then, define \[f(M)=\text{the closure of }\bigcup\tau_n\concat A_n\;.\] (Actually to ensure that $f(M)$ is still nowhere dense, we should select the $\tau_n$ sparse enough.) To see that $f$ is Tukey, suppose that $F$ is in Nwd and that $f(M)\subset F$. Then for all $n$, we have that $A_n\subset F^{-\tau_n}$. It follows that $M\subset\bigcup F^{-\tau_n}$, as desired.

Next, let's attack the problem of finding a $\sigma$-Tukey map from Nwd to Meager. (And no, the identity map does not work!) In similarity with our first construction, we will let \[f(F)=\text{the closure of }\bigcup\tau_n\concat F\;.\] But now we have to be a bit more careful about how we select the $\tau_n$. Instead of fixing the $\tau_n$ in advance, this time they will depend on $F$. As before, we will ensure they are sparse enough that $f(F)$ is meager. But additionally, we can inductively choose them dense enough that the following condition holds:

(*) Whenever $N_\sigma\cap f(F)\neq\emptyset$ there exists $n$ with $\sigma\subset\tau_n$.

Now, to see that $F$ is $\sigma$-Tukey, suppose that $M$ is a meager set such that $f(F)\subset M$. Since $M$ is meager, we can write $M\subset\bigcup A_m$ where $A_m$ are in Nwd. Since $f(F)$ is closed, Baire’s category theorem (!) implies that some $A_m$ must have nonempty interior in the sense of the subspace topology on $f(F)$. This means that there is $\sigma$ such that $N_\sigma\cap f(F)\neq\emptyset$ and $N_\sigma\cap f(F)\subset A_m$. Now appealing to (*) there exists $n$ such that $\sigma\subset\tau_n$. But this implies that $\tau_n\concat F\subset N_\sigma\cap f(F)\subset A_m$, and so $F\subset A_m^{-\tau_n}$.

Putting this together, we conclude that if $f(F)\subset M$ then $F$ is dominated by one of the countable family of $A_m^{-\tau}$ for $m\in\NN$ and $\tau\in2^{<\NN}$, as desired.

Must there be numbers we cannot describe or define? Pointwise definability and the Math Tea argument, Bristol, April 2012

Joel David Hamkins — Thu, 12 Apr 2012 19:13:23 +0000

This is a talk I plan to give to the set theory seminar at the University of Bristol on April 18, 2012.

An old argument, heard at a good math tea, proceeds: “there must be some real numbers that we can neither describe nor define, since there are uncountably many reals, but only countably many definitions.” Does it withstand scrutiny? In this talk, I will discuss the phenomenon of pointwise definable models of set theory, in which every object is definable without parameters. In addition to classical and folklore results on the existence of pointwise definable models of set theory, the main new theorem is that every countable model of ZFC and indeed of GBC has an extension to a model of set theory with the same ordinals, in which every set and class is definable without parameters. This is joint work with Jonas Reitz and David Linetsky, and builds on work of S. Simpson, R. Kossak, J. Schmerl, S. Friedman and A. Enayat.

slides | article

Rectangular square-bracket operation for successor of regulars

Assaf Rinot — Thu, 12 Apr 2012 00:50:44 +0000

Joint work with Stevo Todorcevic.

Extended Abstract: Consider the coloring statement $\lambda^+\nrightarrow[\lambda^+;\lambda^+]^2_{\lambda^+}$ for a given regular cardinal $\lambda$:

In 1990, Shelah proved the above for $\lambda>2^{\aleph_0}$;
In 1991, Shelah proved the above for $\lambda>\aleph_1$;
In 1996, Shelah proved the above for $\lambda=\aleph_1$;
In 2006, Moore proved the above for $\lambda=\aleph_0$.

In this paper, we provide a uniform proof of the fact that $\lambda^+\nrightarrow[\lambda^+;\lambda^+]^2_{\lambda^+}$ holds for every regular cardinal $\lambda$.

Downloads:

Comparing rectangles with squares through rainbow sets

Assaf Rinot — Mon, 09 Apr 2012 04:34:27 +0000

In Todorcevic’s class last week, he proved all the results of Chapter 8 from his Walks on Ordinals book, up to (and including) Theorem 8.1.11. The upshots are as follows:

Every regular infinite cardinal $\theta$ admits a naturally defined function $osc:[\mathcal P(\theta)]^2\rightarrow\omega$;
there exists, again natural, notion of unbounded subfamily of $\mathcal P(\theta)$, and if $\mathcal X$ is an unbounded subfamily of $\mathcal P(\theta)$, then $osc“[\mathcal X]^2=\omega$;
if $\theta=cf(\theta)>\omega$ carries a non-trivial $C$-sequence (i.e., a sequence $\langle C_\alpha\mid\alpha<\theta\rangle$ such that $C_\alpha$ is a club in $\alpha$ for all limit $\alpha<\theta$, and such that for every club $D\subseteq\theta$, there exists an accumulation point $\alpha$ of $D$ such that $D\cap\alpha\nsubseteq C_\beta$ for all $\beta<\theta$), then, roughly speaking, there is a way to convert cofinal sets $A\subseteq\theta$ into sets $A’$ for which $\langle C_\alpha\mid \alpha\in A’\rangle$ forms an unbounded subfamily of $\mathcal P(\theta)$;
building on the previous item, one can prove that every regular uncountable cardinal $\theta$ that carries a non-trivial $C$-sequence, admits a function $o:[\theta]^2\rightarrow\omega$ with the property that $o“[A]^2=\omega$ for every cofinal $A\subseteq\theta$.

I asked the lecturer whether a rectangular version of the above theorems is known. He started by saying that (in general) there is a huge difference between rectangular properties and square properties, and then provided the following:

if $\mathcal X,\mathcal Y$ are unbounded subfamilies of $\mathcal P(\theta)$, then $osc[\mathcal X\times \mathcal Y]$ is co-finite;
it is unknown whether there is an analogous way to convert cofinal subsets $A,B$ of $\theta$ to cofinal $A’,B’$ in way that will yield a function $o:[\theta]^2\rightarrow\omega$ with the property that $o“[A\circledast B]^2=\omega$ for every cofinal $A,B\subseteq\theta$.

Naturally, after proving that every successor of a singular cardinal admits a function that transforms rectangles into squares, I’ve been periodically looking for a mathematical evidence for this well-agreed “huge” difference between the two forms. Of course, there is this meta-evidence that it took two decades between Todorcevic’s theorem that $\omega_1\nrightarrow[\omega_1]^2_{\omega_1}$ holds in ZFC, to Moore’s theorem that $\omega_1\nrightarrow[\omega_1;\omega_1]^2_{\omega_1}$ holds in ZFC, yet, it was only today that I finally found what I was looking for.

In a plenary lecture at the Logic Colloquium 2008, Lajos Soukup mentions a concept (and results) that allows to contrast the rectangular version with the square version, as follows.

Theorem 1 (Erdos-Hajnal, 1978). If $f$ witnesses $\omega_1\nrightarrow[\omega_1;\omega_1]^2_{\kappa}$, then to any coloring $d:[\omega]^2\rightarrow\kappa$, there exists a strictly-increasing function $\psi:\omega\rightarrow\omega_1$ such that $$f(\psi(n),\psi(m))=d(n,m),\quad\text{ for all }n<m<\omega.$$

(so $f$ is universal for countable colorings $d:[\omega]^2\rightarrow\kappa$)

Theorem 2 (Shelah, 1975). CH entails the existence of a coloring $f$ that witnesses $\omega_1\nrightarrow[\omega_1]^2_\omega$, while $f\restriction[B]^2$ is not injective for every subset $B\subseteq\omega_1$ of size $>2$.

(so $f$ does not even embed injections of the form $d:[3]^2\rightarrow 3$)

Now, this is a huge difference indeed.

Before turning to the proofs, let us introduce the notion of a rainbow set. Given a coloring $f:[\lambda]^2\rightarrow\kappa$, we say that $X\subseteq\lambda$ is $f$-rainbow if $f\restriction[X]^2$ is injective. Then:

Trivially, every coloring $f:[\lambda]^2\rightarrow\kappa$ with $\lambda\ge 2$ admits an $f$-rainbow set of size $2$.
Theorem 2 asserts that CH entails the existence of a coloring witnessing $\omega_1\nrightarrow[\omega_1]^2_{\omega}$ that does not admit a rainbow set of size $3$. Moreover, Shelah proved that $\diamondsuit(E^{\lambda^+}_\lambda)$ entails the existence of a coloring witnessing $\lambda^+\nrightarrow[\lambda^+]^2_{\lambda^+}$ that does not admit a rainbow set of size $3$. According to Hajnal, it is unknown whether these hypotheses are necessary.
By Theorem 1, every coloring witnessing $\omega_1\nrightarrow[\omega_1;\omega_1]^2_{\omega}$ admits an infinite rainbow set. Hajnal proved that the same conclusion holds for coloring witnessing $\omega_1\nrightarrow[\omega_1,\omega_1]^2_{\omega}$.
In his paper, Soukup points out that CH entails a coloring witnessing $\omega_1\nrightarrow[\omega;\omega_1]^2_{\omega_1}$ that does not admit an uncountable rainbow set. We noticed that any function witnessing $\omega_1\nrightarrow[\omega_1]^2_{\omega_1}$ does not admit an uncountable rainbow set. Either of the results shows that Theorem 1 cannot be improved to get universality for uncountable colorings.

Proof of Theorem 1. Suppose that $f$ witnesses $\omega_1\nrightarrow[\omega_1;\omega_1]^2_{\kappa}$ (or just $\omega_1\nrightarrow[\text{stat}(\omega_1);\text{stat}(\omega_1)]^2_\kappa$).

Subclaim. For every sequence $\langle S_n\mid n<\omega\rangle$ of stationary subsets of $\omega_1$, and every function $g:\omega\rightarrow\kappa$, there exists a sequence $\langle T_n\mid n<\omega\rangle$ such that:

$T_0\subseteq S_0$ is a singleton;
$T_n\subseteq S_n$ is stationary whenever $0<n<\omega$;
$f[T_0\circledast T_n]=\{g(n)\}$ whenever $0<n<\omega$;
$\max(T_0)<\min(T_n)$ whenever $0<n<\omega$.

Proof of subclaim. Suppose not, as witnessed by $\langle S_n\mid n<\omega\rangle$, and $g$. Then, for every $\alpha\in S_0$, there exists some positive $n<\omega$ such that $f[\{\alpha\}\circledast T_n]\neq\{g(n)\}$ for every stationary $T_n\subseteq S_n$. It follows that there exists some positive $m<\omega$, and a stationary $T\subseteq S_0$ such that $f[\{\alpha\}\circledast T_m]\neq\{g(m)\}$ for every $\alpha\in T$, and every stationary $T_m\subseteq S_m$. Pick clubs $\langle C_\alpha\mid \alpha<\omega_1\rangle$ such that $C_\alpha$ is disjoint from $\{ \beta\in S_m\mid f(\alpha,\beta)=g(m)\}$ for all $\alpha\in T$. Let $C=\Delta_{\alpha<\omega_1}C_\alpha$ be the diagonal intersection, and $T_m:=C\cap S_m$. Then for every $(\alpha,\beta)\in T\circledast T_m$, we get that $\beta\in C_\alpha$, and hence $f(\alpha,\beta)\neq g(m)$, contradicting the fact that $f[T\circledast T_m]=\kappa$. $\square$

Now, suppose that we are given $d:[\omega]^2\rightarrow\kappa$. For all $i<\omega$, let $g_i:\omega\rightarrow\kappa$ be a function satisfying $g_i(n)=d(i,n)$ whenever $i<n<\omega$. By a recursive application of the preceding subclaim, we may then find a matrix $\langle S_{i,n}\mid i<\omega,n<\omega\rangle$ such that:

$S_{0,n}=\omega_1$ for all $n<\omega$;
$S_{i+1,i}\subseteq S_{i,i}$ is a singleton for all $i<\omega$;
$S_{i+1,i+n}\subseteq S_{i,i+n}$ is stationary whenever $i<\omega$ and $0<n<\omega$;
$f[S_{i+1,i}\circledast S_{i+1,i+n}]=\{g_i(n)\}$, whenever $i<\omega$ and $0<n<\omega$;
$\max(S_{i+1,i})<\min(S_{i+1,i+n})$, whenever $i<\omega$ and $0<n<\omega$.

Let $\psi:\omega\rightarrow\omega_1$ be the function satisfying $S_{i+1,i}=\{\psi(i)\}$ for all $i<\omega$. Then, $\psi$ is strictly increasing, and for every $i<n<\omega$, we have $\psi(i)\in S_{i+1,i}$ and $\psi(n)\in S_{i+1,n}$, and hence $$f(\psi(i),\psi(n))=g_i(n)=d(i,n). \square$$

Proof of Theorem 2. For all distinct $\eta,\rho\in{}^\omega2$, let $$\Delta(\eta,\rho):=\min\{ n<\omega\mid \eta(n)\neq\rho(n)\}.$$ Let $h:\omega\rightarrow\omega$ be a surjection such that $h^{-1}\{n\}$ is infinite for all $n<\omega$, and then define $f:[{}^\omega2]^2\rightarrow\omega$ by letting for all distinct $\eta,\rho\in{}^\omega2$: $$f(\eta,\rho):=h(\Delta(\eta,\rho)).$$

It is clear that for every $B\subseteq{}^\omega2$ of size $>2$, the function $f\restriction[B]^2$ is not injective, hence, our main goal is to find an uncountable $X\subseteq{}^\omega2$ for which $f\restriction[X]^2$ witnesses $\omega_1\nrightarrow[\omega_1]^2_{\omega}$.

Here goes. By CH, let $\{ A_i \mid i<\omega_1\}$ be some enumeration of $[{}^\omega2]^{\aleph_0}$. We now define a family $\{ \eta_i\mid i<\omega_1\}\subseteq{}^\omega2$ by recursion on $i<\omega_1$.
For the base case, we let $\eta_0$ be an arbitrary element of ${}^\omega2$.
Now, suppose that $i<\omega_1$ is nonzero, and that $\{ \eta_j\mid j<i\}$ has already been defined. Let

$\{ A_k^{i}\mid k<\omega\}$ be some enumeration of $\{ A_j \mid j<i\}$, and
$\{ \eta_k^{i}\mid k<\omega\}$ be some enumeration of $\{ \eta_j \mid j<i\}$.

$\eta_i$ will be defined as the limit of an increasing chain $\{ \rho_{i,k}\mid k<\omega\}\subseteq{}^{<\omega}2$ which will be set by recursion on $k<\omega$. Of course, we commence with letting $\rho_{i,0}:=\emptyset$.
Next, if $k<\omega$ and $\rho_{i,k}$ has been defined, we consider two cases:

$\rho_{i,k}\subseteq\sigma$ for some $\sigma\in A_{h(k)}^i$. In this case, we fix such $\sigma$ and find a large enough $l<\omega$ such that $|\rho_{i,k}|<l$ and $h(l)=|\{ t<k\mid h(t)=h(k)\}|$. We then let $$\rho_{i,k+1}:=(\sigma\restriction l){}^\frown\langle 1-\sigma(l)\rangle{}^\frown\langle 1-\eta^i_k(l+1)\rangle.$$
otherwise. In this case, we let $l:=|\rho_{i,k}|$, and put: $$\rho_{i,k+1}:=\rho_{i,k}{}^\frown\langle 1-\eta^i_k(l)\rangle.$$

Let $\eta_i:=\bigcup_{k<\omega}\rho_{i,k}$. Then, $\eta_i\in{}^\omega2$ and for all $k<\omega$, $\rho_{i,k+1}\subseteq\eta_i$, while $\rho_{i,k+1}\not\subseteq\eta^i_k$. In particular, $\eta_i\not\in\{\eta_j \mid j<i\}$. It follows that $X:=\{ \eta_i\mid i<\omega_1\}$ is uncountable.

Subclaim. $f\restriction[X]^2$ witnesses $\omega_1\nrightarrow[\omega_1]^2_{\omega}$.
Proof. Suppose that $A\subseteq X$ is uncountable, and $n<\omega$. We shall prove that $n\in f“[A]^2$. Fix a countable elementary submodel $M\prec H_{\omega_2}$ with $A\in M$. Let $j<\omega_1$ be such that $A_j=A\cap M$. Since $A$ is uncountable, pick a large enough $i\in(j,\omega_1)$ such that $\eta_i\in A\setminus M$. Let $m<\omega$ be such that $A_j=A^i_m$. Let $K:=\{ k<\omega\mid h(k)=m\}$. Let $k$ be the $n_{th}$ element of $K$. Then:

$h(k)=m$;
$|\{ t<k\mid h(t)=h(k)\}|=n$;
$A^i_{h(k)}=A\cap M$.

As $\rho_{i,k}\in{}^{<\omega}2\subseteq M$ and $A\in M$, we get that $B:=\{ \sigma\in A\mid \rho_{i,k}\subseteq \sigma\}$ is in $M$. Since $\eta_i$ witnesses that $B$ is non-empty, we infer the existence of some $\sigma\in A\cap M$ such that $\rho_{i,k}\subseteq\sigma$. So, $\rho_{i,k+1}$ has been defined according to case 1, and there exists some $\sigma\in A\cap M$ and $l<\omega$ such that

$h(l)=|\{ t<k\mid h(t)=h(k)\}|=n$, and
$\eta_i\restriction(l+1)=(\sigma\restriction l){}^\frown\langle 1-\sigma(l)\rangle$.

It follows that $\Delta(\sigma,\eta_i)=l$, and hence $f(\sigma,\eta_i)=h(l)=n$. $\square$ $\square$

Proof of Soukup’s theorem. Recall the original construction of Erdos-Hajnal-Milner from CH of a function witnessing $\omega_1\nrightarrow[\omega;\omega_1]^2_{\omega_1}$. Let $\{ A_\gamma\mid \gamma<\omega_1\}$ be some enumeration of $[\omega_1]^{\aleph_0}$. For all infinite $\beta<\omega_1$ do the following:

fix a surjection $\psi_\beta:\omega\rightarrow\beta$ such that $\psi_\beta^{-1}\{\gamma\}$ is infinite for all $\gamma<\beta$;
pick an injection $\phi_\beta:\omega\rightarrow\omega_1$ such that $\phi_\beta(i)\in A_{\psi_\beta(i)}$ for all $i<\omega$. This can be done recursively, where at step $i<\omega$, we utilize the fact that $$|\phi_\beta[i]|=|i|<\aleph_0=|A_{\psi_\beta(i)}|.$$
for all $\gamma<\beta$, denote $A_{\gamma,\beta}:=\{ \phi_\beta(i)\mid \psi_\beta(i)=\gamma\}$. Then $A_{\gamma,\beta}$ is an infinite subset of $A_\gamma$, and $\gamma_1<\gamma_2<\beta$ implies $A_{\gamma_1,\beta}\cap A_{\gamma_2,\beta}=\emptyset$.
for all $\gamma<\beta$, fix a surjection $g_{\gamma,\beta}:A_{\gamma,\beta}\rightarrow\beta$.

Finally, pick a function $f:[\omega_1]^2\rightarrow\omega_1$ such that for all $\alpha<\beta<\omega_1$:

$$f(\alpha,\beta)=\begin{cases}
0,&\alpha\not\in\bigcup_{\gamma<\beta}A_{\gamma,\beta}\\
g_{\gamma,\beta}(\alpha),&\alpha\in A_{\gamma,\beta}
\end{cases}.$$

Subclaim. $f$ witnesses $\omega_1\nrightarrow[\omega;\omega_1]^2_{\omega_1}$.
Proof. Suppose that $Y$ is an infinite countable subset of $\omega_1$, $Z$ is an uncountable subset of $\omega_1$, and $\delta<\omega_1$. We shall show that $\delta\in f[Y\circledast Z]$.

Fix $\gamma<\omega_1$ such that $A_\gamma=Y$. Pick a large enough $\beta\in Z$ such that $\beta>\max\{\gamma,\delta,\sup(Y)\}$.

Since $\delta<\beta$ and $g_{\gamma,\beta}:A_{\gamma,\beta}\rightarrow\beta$ is surjective, we may now pick $\alpha\in A_{\gamma,\beta}$
such that $g_{\gamma,\beta}(\alpha)=\delta$. Then $\alpha\in A_{\gamma,\beta}\subseteq Y\cap\beta$ and $$f(\alpha,\beta)=g_{\gamma,\beta}(\alpha)=\delta. \square$$

It now follows from the upcoming observation that the function just constructed does not admit an uncountable rainbow set. $\square$

Observation. If $f$ witnesses $\kappa\nrightarrow[\kappa]^2_\theta$, then $f$ does not admit a rainbow set of size $\min\{\kappa,\theta^+\}$.
Proof. Given a set $X$ of size $\kappa$, decompose $X$ as $X_0\uplus X_1$, such that $X_i$ is of size $\kappa$, for all $i<2$. Then $f“[X_0]^2=\theta$ and $f“[X_1]^2=\theta$. In particular $f\restriction [X_0\cup X_1]^2$ is not injective. Given a set $X$ of size $\theta^+$, it is trivial that $f\restriction[X]^2:[X]^2\rightarrow\theta$ is not injective. $\square$

Rapid idempotent ultrafilters

Peter Krautzberger — Sun, 08 Apr 2012 17:58:50 +0000

Welcome back to the second (and final) part on why strongly summable ultrafilters are rapid.

Why the new title? Well, first, I needed another one (I’ve had too many posts with “Part X” in them, I think). Second, after …

A course in infinitary computability, Fall 2012, CUNY Graduate Center, CSC 85020

Joel David Hamkins — Sun, 08 Apr 2012 09:57:26 +0000

In Fall 2012 I will teach a graduate course on infinitary computability theory in the Computer Science program at the CUNY Graduate Center. This course will be aimed at graduate students in computer science and mathematics who are interested in infinitary computational processes.

Infinitary computability, CSC 85020, Mondays, 9:30 – 11:30 am
CS course listings | this listing

This course will explore all the various infinitary theories of computability, including infinite time Turing machines, Blum-Shub-Smale computability, Bucci automata, ordinal register machines and others. The focus will be on introducing the computability models and comparing them to each other and to standard concepts of computability. In the early part of the course, we shall review the standard finitary computational models, before investigating the infinitary supertask analogues.

I will post further information here about readings, homework and so on, so this page will become a home for the course. (Note comment feature below for questions.)

Students wishing to prepare for the course should review their understanding of Turing machines and the other theoretical machine models of computation.

Some of my articles on infinitary computability

Fun and paradox with large numbers, logic and infinity, Philadelphia 2012

Joel David Hamkins — Fri, 06 Apr 2012 11:21:27 +0000

This is a fun talk I will give at Temple University for the mathematics undergraduates in the Senior Problem Solving forum. We’ll be exploring some of the best puzzles and paradoxes I know of that arise with large numbers and infinity. Many of these paradoxes are connected with deep issues surrounding the nature of mathematical truth, and my intention is to convey some of that depth, while still being accessible and entertaining.

Abstract: Are there some real numbers that in principle cannot be described? What is the largest natural number that can be written or described in ordinary type on a 3×5 index card? Which is bigger, a googol-bang-plex or a googol-plex-bang? Is every natural number interesting? Is every true statement provable? Does every mathematical problem ultimately reduce to a computational procedure? Is every sentence either true or false or neither true nor false? Can one complete a task involving infinitely many steps? We will explore these and many other puzzles and paradoxes involving large numbers, logic and infinity, and along the way, learn some interesting mathematics.

What happens when one iteratively computes the automorphism group of a group? Temple University, Philadelphia 2012

Joel David Hamkins — Fri, 06 Apr 2012 11:01:43 +0000

This is a talk I shall give for the Mathematics Colloquium at Temple University, April 23, 2012.

The automorphism tower of a group is obtained by computing its automorphism group, the automorphism group of that group, and so on, iterating transfinitely. The question, known as the automorphism tower problem, is whether the tower ever terminates, whether there is eventually a fixed point, a group that is isomorphic to its automorphism group by the natural map. Wielandt (1939) proved the classical result that the automorphism tower of any finite centerless group terminates in finitely many steps. This was successively generalized to larger and larger collections of groups until Thomas (1985) proved that every centerless group has a terminating automorphism tower. Building on this, I proved (1997) that every group has a terminating automorphism tower. After giving an account of this theorem, I will give an overview of work with Simon Thomas and newer work with Gunter Fuchs and work of Philipp Lücke, which reveal a set-theoretic essence for the automorphism tower of a group: the very same group can have wildly different towers in different models of set theory.

slides | list of my articles on automorphism towers | announcement poster

The automorphism tower problem for groups, Bristol 2012

Joel David Hamkins — Fri, 06 Apr 2012 10:39:59 +0000

Isaac Newton 20th Anniversary Lecture. This is a talk I shall give at the University of Bristol, School of Mathematics, April 17, 2012, at the invitation of Philip Welch.

The automorphism tower of a group is obtained by computing its automorphism group, the automorphism group of that group, and so on, iterating transfinitely. The question, known as the automorphism tower problem, is whether the tower ever terminates, whether there is eventually a fixed point, a group that is isomorphic to its automorphism group by the natural map. Wielandt (1939) proved the classical result that the automorphism tower of any finite centerless group terminates in finitely many steps. This was successively generalized to larger and larger collections of groups until Thomas (1985) proved that every centerless group has a terminating automorphism tower. Building on this, I proved (1997) that every group has a terminating automorphism tower. After giving an account of this theorem, I will give an overview of my work with Simon Thomas, as well as newer work with Gunter Fuchs and work of Philipp Lücke, which reveal a set-theoretic essence for the automorphism tower of a group: the very same group can have wildly different towers in different models of set theory.

slides | list of my articles on automorphism towers | abstract at Bristol

Quiz on public peer review

Carl Mummert — Thu, 05 Apr 2012 00:16:07 +0000

I have been required to complete a “responsible conduct of research” training module by the research office at my school. The reason I am commenting is that I was asked to answer the following question “true” or “false”. This is … Continue reading →

Another Combinatorial Result

Micheal Pawliuk — Wed, 04 Apr 2012 16:45:14 +0000

Here is Chris Eagle’s presentation from Stevo Todorcevic’s class “Combinatorial Set Theory”.

From the abstract:

We prove that MA + $\mathfrak{c} = \aleph_2$ implies $\mathfrak{c} \not\rightarrow (\mathfrak{c}, \omega+2)^2$ . The exposition is based on hand-written notes provided by S. Todorcevic. The result itself is due to R. Laver.

This is the analogous result to “MA implies (NonSpecial Tree) $\not\rightarrow$ (NonSpecial Tree, $\omega+2)^2$”, which I explained here.

One day in Colorado or Strongly summable ultrafilters are rapid

Peter Krautzberger — Tue, 03 Apr 2012 01:21:27 +0000

Picking up from the prelude about micro-contributions, let me tell you the story of a small result I would like to share with you. (Before you frantically scroll down to find the proof, it’s not really here but will …

2012 North American Annual Meeting of the ASL

Assaf Rinot — Fri, 30 Mar 2012 01:07:38 +0000

I gave a special session talk at the ASL 2012 North American Annual Meeting (Madison, March 31–April 3, 2012).

Talk Title: The extent of the failure of Ramsey’s theorem at successor cardinals.

Extended abstract: Ramsey’s theorem asserts that for every coloring $c:[\omega]^2\rightarrow2$, there exists an infinite subset $H\subseteq\omega$ such that $c\restriction [H]^2$ is constant. At the early 1930′s, Sierpinski showed that a generalization of Ramsey’s theorem must fail for successor cardinals, and ever since the extent of this failure was studied extensively by many set-theorists, including, Erdos, Eisworth, Hajnal, Moore, Shelah, and Todorcevic.

In this talk, we shall show that Shelah’s notion of strong coloring $\text{Pr}_0(\lambda^+,\lambda^+,\lambda^+,\omega)$ coincides with the most basic concept considered already by Erdos and his collaborators: $\lambda^+\not\rightarrow[\lambda^+]^2_{\lambda^+}$. More specifically, we shall discuss the following ZFC result.
Theorem. The following are equivalent for every uncountable cardinal $\lambda$:
(1) There exists a coloring $c:[\lambda^+]^2\rightarrow\lambda^+$ such that for every
(-) color $\gamma<\lambda^+$, and every
(-) subset $A\subseteq\lambda^+$ of size $\lambda^+$,
there exist $\alpha,\beta\in A$ with $\alpha<\beta$ such that $$c(\alpha,\beta)=\gamma.$$
(2) There exists a coloring $c:[\lambda^+]^2\rightarrow\lambda^+$ such that for every
(-) coloring $g:n\times n\rightarrow\lambda^+$ (here $n$ is a positive integer), and every
(-) family $\mathcal A\subseteq[\lambda^+]^n$ of size of $\lambda^+$ of mutually disjoint sets,
there exist $a,b\in\mathcal A$ with $\max(a)<\min(b)$ such that $$c(a_i,b_j)=g(i,j)\text{ for all }i,j<n.$$

(here, $a_i$ denotes the $i_{th}$-element of $a$, and $b_j$ denotes the $j_{th}$-element of $b$.)

Downloads:

Julia sets and the Mandelbrot set

Victoria Gitman — Thu, 29 Mar 2012 19:42:12 +0000

This is a talk at the City Tech Math Club, April 19, 2012.

I wanted to learn about Julia sets and the Mandelbrot set ever since my undergraduate course in dynamical systems never quite got there. The material was the culmination of Robert Devaney’s “A first course in dynamical systems: theory and experiment”, which I have looked at longingly for many years, but what with being a mathematician and all, never got a chance (um, excuse) to complete. So naturally I jumped at the opportunity to give a talk about it at our math club and plowed right through the textbook. Of course, anyone who is familiar with the definition of Julia sets and the Mandelbrot set knows that you don’t need to know much of anything about dynamical systems to make sense of it. Still, I think something is lost by not putting these notions in the larger context of dynamical systems concepts such as orbits, attracting cycles, and chaotic behavior. The same must be said about the complex numbers. So I structured the talk in four parts. The first part is an infomercial for the material consisting of the most gorgeous images available on the web. The second is a crash course in dynamical systems. The third is a review of the complex numbers that includes history, motivation, and the graphical representation. The actual culprits of the talk make up the final part.

Next, I copied and pasted my usual Beamer code and prepared to make the slides. Beamer, the LaTeX package for making slides, has served me well over the years. It is a pretty perfect tool for presenting definitions, theorems, and proofs about forcing and large cardinals. However, as I quickly discovered, it is rather badly suited to a visually stimulating presentation on Julia sets. It is difficult to impossible to control where an image gets placed on the slide and forget about precise resizing. If you happen to want a large enough image, it will take up the entire slide, leaving no room for explanatory text. The beautiful GIF animations of the Mandelbrot set are not supported. After a wasted morning of trying to make it all work, I started to take a stock of my limited options. PowerPoint works great for visuals but typing math is a pain and the result looks ugly to boot. Beamer was having an adverse affect on my nerves. Then a totally crazy answer hit me: HTML. My slides are going to webpages!

Pros:
1) I know enough HTML, CSS and Javascript to pull it off.
2) I can create my dream layout.
3) I can put everything related to a particular concept on a single slide (aka webpage), no more running out of room.
4) I can have beautiful math compiled from LaTeX code courtesy of the marvelous MathJax.

Cons:
1) A slide won’t necessarily fit on the screen, so scrolling will need to take place.

In the balance, it clearly looked like a winner!

Judge for yourself here.

Prelude to a small experiment

Peter Krautzberger — Tue, 27 Mar 2012 01:49:01 +0000

I haven’t posted anything in almost two months — life happened. It’s still a bit chaotic and maybe I’ll write about it when things calm down. For now, I’m back from a productive trip to Toronto where, among other things, …

A talk on Tukey maps

Samuel Coskey — Tue, 27 Mar 2012 01:25:07 +0000

I will be speaking this Thursday, March 29 at the University of Toronto Student Set Theory and Topology Seminar about Tukey maps between partial orders.

There are many reasons to study Tukey reductions. For instance, many important cardinal invariants can be realized as the cofinality of a partial order. And here, a Tukey reduction between the partial orders corresponds to an inequality between the cardinal invariants.

In the talk I will explain this in more detail, then I will sketch some work of Slawek and Stevo which gives ‘automatic definability’ of Tukey reductions between certain very special partial orders.

PS: Of course, numerous applications to forcing, C*-algebras, and Dominion will be spread homogeneously throughout the talk.

Bumper sticker

François G. Dorais — Sat, 24 Mar 2012 16:38:43 +0000

An amazing idea for a bumper sticker from xkcd: Of course, this has many implications for people driving behind you…

The hierarchy of equivalence relations on the natural numbers under computable reducibility, Cambridge, March 2012

Joel David Hamkins — Fri, 23 Mar 2012 20:42:16 +0000

This is a talk I shall give at the workshop on Logical Approaches to Barriers in Complexity II, March 26-30, 2012, a part of the program Semantics and Syntax: A Legacy of Alan Turing at the Isaac Newton Institute for Mathematical Sciences in Cambridge, where I am currently visiting.

Many of the naturally arising equivalence relations in mathematics, such as isomorphism relations on various types of countable structures, turn out to be equivalence relations on a standard Borel space, and these relations form an intensely studied hierarchy under Borel reducibility. The topic of this talk is to introduce and explore the computable analogue of this robust theory, by studying the corresponding hierarchy of equivalence relations on the natural numbers under computable reducibility. Specifically, one relation $E$ is computably reducible to another $F$ , if there is a unary computable function $f$ such that $x\mathrel{E}y$ if and only if $f(x)\mathrel{F}f(y)$. This gives rise to a very different hierarchy than the Turing degrees on such relations, since it is connected with the difficulty of the corresponding classification problems, rather than with the difficulty of computing the relations themselves. The theory is well suited for an analysis of equivalence relations on classes of c.e. structures, a rich context with many natural examples, such as the isomorphism relation on c.e. graphs or on computably presented groups. An abundance of open questions remain, and the subject is an attractive mix of methods from mathematical logic, computability, set theory, particularly descriptive set theory, and the rest of mathematics, subjects in which many of the equivalence relations arise. This is joint work with Sam Coskey and Russell Miller.

Slides | Slides (longer version, from a similar talk) | Article | Conference abstract | Video

Infinite chess: the mate-in-n problem is decidable and the omega-one of chess, Cambridge, March 2012

Joel David Hamkins — Thu, 22 Mar 2012 21:56:59 +0000

I have just taken up a visiting fellow position at the Isaac Newton Institute for mathematical sciences in Cambridge, UK, where I am participating in the program Syntax and Semantics: the legacy of Alan Turing. I was asked to give a brief introduction to some of my current work, and I chose to speak about infinite chess.

Nevertheless, in joint work with Dan Brumleve and Philipp Schlicht, confirming a conjecture of myself and C. D. A. Evans, we establish that the mate-in-$n$ problem of infinite chess is computably decidable, uniformly in the position and in $n$. Furthermore, there is a computable strategy for optimal play from such mate-in-$n$ positions. The proof proceeds by showing that the mate-in-$n$ problem is expressible in what we call the first-order structure of chess, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable. Unfortunately, this resolution of the mate-in-n problem does not appear to settle the decidability of the more general winning-position problem, the problem of determining whether a designated player has a winning strategy from a given position, since a position may admit a winning strategy without any bound on the number of moves required. This issue is connected with transfinite game values in infinite chess, and the exact value of the omega one of chess $\omega_1^{\rm chess}$ is not known. I will also discuss recent joint work with C. D. A. Evans and W. Hugh Woodin showing that the omega one of infinite positions in three-dimensional infinite chess is true $\omega_1$: every countable ordinal is realized as the game value of such a position.

article | slides | streaming video | program of abstracts

Is the dream solution of the continuum hypothesis attainable?

Joel David Hamkins — Tue, 20 Mar 2012 04:14:25 +0000

J. D. Hamkins, “Is the dream solution of the continuum hypothesis attainable?,” , pp. 1-10. (submitted)

Citation arχiv

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AUTHOR = {Joel David Hamkins},
TITLE = {Is the dream solution of the continuum hypothesis attainable?},
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pages = {1--10},
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Many set theorists yearn for a definitive solution of the continuum problem, what I call a dream solution, one by which we settle the continuum hypothesis (CH) on the basis of a new fundamental principle of set theory, a missing axiom, widely regarded as true, which determines the truth value of CH. In an earlier article, I have described the dream solution template as proceeding in two steps: first, one introduces the new set-theoretic principle, considered obviously true for sets in the same way that many mathematicians find the axiom of choice or the axiom of replacement to be true; and second, one proves the CH or its negation from this new axiom and the other axioms of set theory. Such a situation would resemble Zermelo’s proof of the ponderous well-order principle on the basis of the comparatively natural axiom of choice and the other Zermelo axioms. If achieved, a dream solution to the continuum problem would be remarkable, a cause for celebration.

In this article, however, I argue that a dream solution of CH has become impossible to achieve. Specifically, what I claim is that our extensive experience in the set-theoretic worlds in which CH is true and others in which CH is false prevents us from looking upon any statement settling CH as being obviously true. We simply have had too much experience by now with the contrary situation. Even if set theorists initially find a proposed new principle to be a natural, obvious truth, nevertheless once it is learned that the principle settles CH, then this preliminary judgement will evaporate in the face of deep experience with the contrary, and set-theorists will look upon the statement merely as an intriguing generalization or curious formulation of CH or $\neg$CH, rather than as a new fundamental truth. In short, success in the second step of the dream solution will inevitably undermine success in the first step.

This article is based upon an argument I gave during the course of a three-lecture tutorial on set-theoretic geology at the summer school Set Theory and Higher-Order Logic: Foundational Issues and Mathematical Development, at the University of London, Birkbeck in August 2011. Much of the article is adapted from and expands upon the corresponding section of material in my article The set-theoretic multiverse.

The math house rules

Samuel Coskey — Mon, 19 Mar 2012 15:30:11 +0000

Everyone is familiar with the cider house rules (shown below). But not everyone is familiar with Avital’s rules of informal mathematics discussions.

Here are the original cider house rules, as layed down in the academy award winning Michael Caine film:

Please don’t smoke in bed.
Please don’t go up to the roof to eat your lunch.
Please—even if you are very hot—do not go up to the roof to sleep.
There should be no going up on the roof at night.

My favorite part is that after rule (2) is read, somebody shouts out, “That’s the best place to eat lunch!”

A couple of weeks ago, I participated in a math hangout led by Avital Oliver. This hangout was very similar, if much better organized, to the ones I have organized recently. And this wasn’t surprising: it was Avital who gave me the idea in the first place.

The hangout begain with a set of rules. Before he stated them, Avital told us his key guiding principle: the goal is not to find answers, but rather to have a mathematical experience. With that in mind, here are my formulation of his rules:

You are not forced to work, you may observe.
Please do not give answers derived from prior knowledge.
Please, even if you are very anxious, ask privately whether your comment is appropriate.
Maybe, after some amount of time has passed, there will be a period after which you can use prior knowledge.

What do you think of these rules? What additional rules would you impose?

Possibly true. Necessarily funny.

François G. Dorais — Sun, 18 Mar 2012 13:38:42 +0000

This is the tagline for fauxphilnews, a hilarious philosophy blog created by Ben Bronner. My favorite entry so far is Saul Kripke’s resignation after faking results of thought experiments. Check it out…

Convergence of ideas

François G. Dorais — Sun, 11 Mar 2012 02:37:30 +0000

As a moderator on MathOverflow, I see a lot of interesting interactions between mathematicians. The occasional dramatic situations get discussed profusely by community but very few take the time talk about the pleasant exchanges they have had. Here is one that caught my attention today because it illustrates how it is not uncommon for two [...]

Pure logic

Assaf Rinot — Fri, 09 Mar 2012 02:07:20 +0000

While traveling downtown today, I came across a sign near a local church, with a quotation of Saint-Exupéry:

“Pure logic is the ruin of the spirit”? So sad.

Facts about the Urysohn Space – Some useful, some cool

Micheal Pawliuk — Wed, 07 Mar 2012 00:44:05 +0000

(This is almost verbatim the talk I gave recently (Feb 23, 2012) at the Toronto Student Set Theory and Topology Seminar. I will be giving this talk again on April 5, 2012)

I have been working on a problem involving the Urysohn space recently, and I figured that I should fill people in with the basic facts and techniques involved in this space. I will give some useful facts, a key technique and 3 cool facts. First, the definition!

Definition: A metric space $U$ has the Urysohn property if

$U$ is complete and separable

$U$ contains every separable metric space as an isometric copy.

$U$ is ultrahomogeneous in the sense that if $A,B$ are finite, isometric subspaces of $U$ then there is an isomorphism of $U$ that takes $A$ to $B$.

You might already know a space that satisfies the first two properties – The Hilbert cube $[0,1]^\omega$ or $C[0,1]$ the continuous functions from $[0,1]$ to $[0,1]$. However, these spaces are not ultrahomogeneous. Should a Urysohn space even exist? It does, but the construction isn’t particularly illuminating so I will skip it.

Usually, ultrahomogeneity and universality are used in the following way:

Fact 1: A Urysohn space has the “1 point extension property”. That is, if $A$ is a finite subset of $U$ and $A\cup\{a\}$ is an arbitrary metric extension of $A$, then there is a point $z \in U$ such that $A\cup \{a\} \cong A \cup\{z\}$ with $A$ getting mapped to $A$ as the identity.

proof. Let $A$ be a finite subset of $U$ and let $A \cup \{c\}$ be a one point metric extension of $A$. Well, $A \cup \{c\}$ is a separable metric space, so find a copy of it in $U$, call it $A’ \cup \{c’\}$. Now use ultrahomogeneity of $U$ to map $A’$ to $A$, by an isometry $\phi$. Then $A \cup \{\phi(c’)\}$ is contained in $U$ and is isometric to $A \cup \{c\}$.[QED]

1 Point extension property

It turns out that the 1 point extension property is a very strong property. This is kind of expected because a space with the 1 point extension property contains a copy of every finite metric space. That is pretty big already!

Fact 2: For a complete, separable metric space $U$, TFAE:

$U$ has the 1pt extension property

$U$ is a Urysohn space

proof. We have already seen the converse, so let us assume that $U$ has the 1pt extension property and we will show that $U$ is universal with respect to separable metric spaces and that $U$ is ultrahomogeneous.

Universal: This is an easy induction. Let $Q = \{q_n : n<\omega\}$ be a dense subset of a metric space. Since $U$ is complete it will be enough to show that $Q \sse U$ (as an isometric copy of course!)

Any finite collection $\{q_i : i<N\}$ is contained (isometrically) in $U$ (by using the 1pt extension property $N$ times) and one more application gives that $\{q_i : i<N\} \cup \{q_N\}$ is contained (isometrically) in $U$.

Since at each stage we are extending, the union of all of these finite copies will be the desired copy of $Q$.

Ultrahomogeneous: For no cost we can actually show a more general fact that illustrates a helpful technique. (This is one of those cases where showing the more general fact more clearly indicates what properties are important.)

Ultrahomogeneity Theorem: Let $X,Y$ be separable, complete metric spaces with the 1pt extension property. Finite isomorphisms extend to the whole space. (i.e. If $A$ is a finite subset of $X$ and $B$ is a finite subset of $Y$ such that $A \cong B$ then there is an isomorphism $f: X \rightarrow Y$ that sends $A$ to $B$.)

This will be a back-and-forth argument. The “back” part will ensure that we have a surjection, and the “forward” will give us a function defined on all of $X$. The proof is by induction, but seeing the first step will be enough to give you the entire idea.

Let $\{x_n : n<\omega\} \sse X$ be countable, dense and let $\{y_n : n<\omega\}$ be countable dense. Since $X,Y$ are complete it will be enough to define a map $f$ on at least $\{x_n : n<\omega\}$ whose image contains $\{y_n : n<\omega\}$ as there is a unique function that extends $f$ to all of $X,Y$.

Let $f_{-1} : A\rightarrow B$ be an isomorphism.

We will define $f_0$ so that:

$f_0$ extends $f_{-1}$
$x_0 \in \dom(f_0)$
$y_0 \in \ran(f_0)$

(Sam’s comment: It might not be possible to map the countable dense subset of $X$ isometrically onto the countable dense subset of $Y$. For example, the rationals cannot be mapped isometrically onto the rationals with $\pi$.)

Now $A \cup \{x_0\} \sse X$ is a finite metric space, and by the 1pt extension property of $Y$ there is a $w_0 \in Y$ such that $B \cup \{w_0\}$ is isometric to $A \cup \{x_0\}$. Obviously there is in fact an isometry $f_{-0.5}: A \cup \{x_0\} \rightarrow B \cup \{w_0\}$ that extends $f_{-1}$.

"Forth" adds $x_0$ to the domain.

So we have extended the domain by one point. Now to extend the range. (The next paragraph is almost a copy of the one above.)

Notice that $B \cup \{w_0\} \cup \{y_0\}$ is a finite metric space, and by the 1pt extension property of $X$ there is a $z_0 \in X$ such that $A \cup \{x_0\}\cup \{z_0\}$ is isometric to $B \cup \{w_0\}\cup \{y_0\}$. Obviously there is in fact an isometry $f_{0}: A \cup \{x_0\}\cup \{z_0\} \rightarrow B \cup \{w_0\} \cup \{y_0\}$ that extends $f_{-0.5}$.

"Back" adds $y_0$ to the range.

So by induction there is a family of isometries $\{f_n : n<\omega\}$ such that

$f_n$ extends $f_{n-1}$
$\{x_i : 0\leq i \leq n\} \in \dom(f_n)$
$\{y_i : 0\leq i \leq n\} \in \ran(f_n)$

Thus $f := \bigcup_{n<\omega} f_n$ is an isometry from $A \cup \{x_n : n<\omega\}\cup \{z_n : n<\omega\}$ onto $B\cup \{w_n : n<\omega\}\cup \{y_n : n<\omega\}$. [QED]

(Sam and Ivan’s comment: It is true a posteriori that $\{w_n : n<\omega\}$ is dense in $Y$ as isometries map dense sets to dense sets. In fact continuous surjections map dense sets to dense sets.)

By letting $X=Y=U$ and $A = B = \emptyset$ in the above theorem we get:

FACT 3: Any two Urysohn spaces are isometric.

And letting $X=Y=U$ we get what some people would call “ultahomogeneity”:

FACT 4: In the Urysohn space $U$, any partial isometry $\phi: A \rightarrow B$, with $A,B \sse U$ finite, then $\phi$ extends to all of $U$.

More on the 1pt extension property

In practise the 1 point extension property isn’t really the right way to describe 1pt extensions. (Wait, what?) It turns out that instead of describing abstract one point metric extensions (which requires using an element outside of the metric space in question) we instead define the distances from a metric space to an outside point. (Stay with me, this will make sense soon.)

Definition: Let $(X,d)$ be a metric space. A function $f: X \rightarrow \R$ is a katetov function if $$\forall x,y \in X, |f(x)-f(y)| \leq d(x,y) \leq f(x) + f(y)$$ We let $E(X)$ be the set of katetov functions on $X$.

At first these seem like weird constraints to put on a function. The first inequality is just saying that $F$ is 1-Lipschitz, the second inequality is a type of triangle inequality (which in particular guarantees that $f$ is non-negative). Intuitively, these functions are meant to represent distances to a point.

Example 1: Let $(X,d)$ be your favourite metric space with $\heartsuit \in X$. Consider $f: X \rightarrow \R$ given by $f(x)=d(x, \heartsuit)$. Notice that $\forall x,y \in X$:

$\|f(x)-f(y)\|$	$=\|d(x,\heartsuit) – d(y,\heartsuit)\|$
	$\leq \|d(x,y)+d(y,\heartsuit) – d(y,\heartsuit)\|$
	$\leq d(x,y)$
	$\leq d(x, \heartsuit)+d(y, \heartsuit)$
	$= f(x)+f(y)$

Example 2: If $f \in E(X), c \geq 0$, then $g(x) = f(x)+c$ is also a katetov function.

Example 3: If $f \in E(X)$ we can get a 1pt extension of $(X, d)$ to $(X \cup \{z\}, d^\prime)$ by letting $d^\prime (x,y)=d(x,y)$ if $x,y \in X$ and $d^\prime (x,z)=f(x)$ if $x \in X$. Notice now that being a katetov function ensures that d’ satisfies the triangle inequality.

FACT 5: If $X \sse U$ is finite, and $f \in E(X)$, then there is a $z \in U$ such that $f(x)=d(x,z)$ for all $x\in X$.

This says that $f$ being a katetov function means that it describes distances to an (abstract) point. (i.e. $f$ describes a 1pt extension). Since $U$ has the 1pt extension property, there is a point $z\in U$ that witnesses this.

Cool Stuff

Enough with all of that abstract stuff, let’s show you some cool stuff.

You are probably wondering about the topological properties of $U$. Well you already know that $U$ is a complete, separable metric space (which is really nice), but what about connectedness?

FACT 6: $U$ has a strong form of path-connectedness, that is, $U$ has geodesics.

Definition: A metric space $X$ has geodesics if for any two points $a,b \in X$ there is a closed interval $I = [x,y]$ and an isometry $f$ from $I$ into $X$ such that $f(x)=a$ and $f(y)=b$. So there is “a line between $a$ and $b$”.

proof. The picture says everything, but for those of you so inclined, here are the details.

Take $a \neq b$, both in $U$. Let $I = [0, d(a,b)]$. Hey, $I$ is a separable metric space, so find an isometric copy $J$ in $U$. Now by ultrahomogeneity of $U$, there is an isometry $f$ of $U$ that maps the end-points of $J$ to $\{a,b\}$. Thus $f$ and $J$ fulfill the requirements of the definition of geodesics. [QED]

See, the picture says everything.

(Dominic’s question: Does $U$ have unique geodesics?)

Fact 7: $U$ has uncountably many geodesics between any two points.

proof. Now we get to use a standard proof technique in Urysohn spaces- We construct katetov functions that code different geodesics.

Fix a single geodesic $I$ between two points $a \neq b$, and let $m$ be the midpoint of $I$. For simplicity, assume that $I$ has length 8.

Consider $f_\delta: \{a,b,m\} \rightarrow \R$ defined by $f_\delta (a)=f_\delta (b)= 4$ and $f_\delta (m)=\delta$, where $0<\delta < 4$.

If we can show that $f_\delta$ is a katetov functions, then it will produce a point $z_\delta \in U$ that is distance $4$ to both $x$ and $y$ (so by the triangle inquality it lies on a geodesic from $x$ to $y$), and each $z_\delta$ is distinct as $d(z_\delta, z_\gamma) \neq 0$ if $\delta \neq \gamma$.

Claim: $f_\delta$ is a katetov function.

(I’ll supress the subscript for readability)

$$|f(a)-f(b)|=|4-4|\leq d(a,b) =8 \leq 4+4 = f(a)+f(b)$$ and $$|f(a)-f(m)|= |4-\delta| \leq 4 \leq d(a,m) \leq 4 + \delta = f(a)+f(m)$$[QED]

Cool Isometric copies of $U$

Now we look at an interesting copy of $U$. For the record, $0 \in U$, which is clear from the construction of $U$ (which I have omitted).

FACT 7: There are isometric copies of $U$ in $U$ with empty interior.

proof. For a finite set $F \sse U$ we define $m(F) = \{x \in U : d(x,a)=d(x,b), \forall a,b \in F\}$.

For example, in $\R^2$, if $F= \{(0,0),(2,0)\}$ then $m(F)= \{(1,y): y \in \R \}$.

Example of m(F)

Claim: For any nonempty finite set $F \sse U$, $m(F) \cong U$.

It is pretty easy to swallow that $m(F)$ is a closed subset of $U$, a complete, separable metric space. Thus $m(F)$ is itself a complete separable metric space and by the ultrahomogeneity theorem it is enough to show that $m(F)$ has the 1pt extension property.

Let $A \sse m(F)$ be finite and let $f: A \rightarrow \R$ be a katetov function.

Let $g: A\cup F \rightarrow \R$ be defined by $g(a)=f(a)$ for $a\in A$ and $g(x)=\inf_{a\in A}\{f(a) + d(a,x)\}$.
[This is called the katetov extension of $f$ to $A \cup F$].

We will show later that this is a katetov function, but if we accept that $g$ is a katetov function then there is a $z \in U$ such that $g(x)=d(z,x)$ for all $x \in A \cup F$.

Since $g(a)=f(a)$ for all $a \in A$, it only remains to see that $z \in m(F)$. This follows since $d(a,x) = d(b,x)$ for all $a,b \in A$, so $d(x,z)=g(x)=g(y)=d(y,z)$ for all $x,y \in F$; the infimum does not depend on $x$ or $y$.
[QED]

This next fact stands in direct opposition of Fact 7:

FACT 8: There are (proper) isometric copies of $U$ in $U$ with non-empty interior.

This will be shown using the following proposition:

Proposition: For $X$ a compact subset of $U$, $M \in \R$, we have $U\setminus \{z \in U : d(z,X )< M\} \cong U$. Thus $U$ is isometric to $U\setminus B(0,1)$.

proof. Note that $\{z \in U : d(z,X )< M\}$ is an open subset of $U$, so $U\setminus \{z \in U : d(z,X )< M\}$ is a complete separable metric space. Thus we only need to show that it has the 1pt extension property.

Let $A \sse U\setminus \{z \in U : d(z,X )< M\}$ be finite, and let $f: A \rightarrow \R$ be a katetov function. Without loss of generality we may assume that $\epsilon := \inf_{a \in A} f(a) >0$.

Let $Y \sse X$ be an $\epsilon$-net. (Every point of $X$ is within $\epsilon$ of some point of $Y$)

Now consider $g: Y \cup A \rightarrow \R$ be defined by $g(x) = \inf_{a \in A} \{f(a)+d(x,a)\}$. Let us accept for now that this is a katetov function.

Then there is a $z \in U$ such that $d(z,x) = g(x)$ for all $x\in Y \cup A$. For $a\in A$ we just get $d(z,a) = f(a)$. For $y \in Y$ we get $$d(z,y) = \inf_{a \in A} \{f(a)+d(x,a)\} > \epsilon + M$$

And so it is clear that $d(z,x)> M$ for all $x \in X$, not just those in $Y$.
[QED]

Katetov Extensions

The only thing we have left to justify is why the functions we defined in Fact 1 and 2 are katetov functions. This comes from a general notion of the “amalgamation of two metric spaces along a compact subspace”, which is mumbo jumbo for gluing two space together.

The idea is that if you have two unrelated spaces that both have an (isometric) copy of the unit circle (or your favourite compact metric space) in them, then you can define a bigger space that contains your original spaces, glued together at the circle.

Amalgamation of A and B along X.

Showing that there is such a space will prove that the functions $g: Y \cup A \rightarrow \R$ be defined by $g(x) = \inf_{a \in A} \{f(a)+d(x,a)\}$ really are katetov functions (where $f: A \rightarrow \R$ is katetov).

FACT 9: Amalgamation is possible.

proof. Let $(A, d_A)$ and $(B, d_B)$ be metric spaces both having an isometric copy of $X$, a compact metric space. Let $Z = A \cup B$, but identify the copies of $X$. Now we need to define a metric $d$ on $Z$ that agrees with $A$ and $B$.

So let $d(a_1, a_2)= d_A(a_1, a_2)$ if $a_1, a_2 \in A$. Similarly, let $d(b_1, b_2)= d_B (b_1, b_2)$ if $b_1, b_2 \in B$.

Now we only need to deal with the case where the points are in different spaces.

Let $d(a,b) = \inf_{x\in X} (d_A (a,x)+d_B (x,b))$. So it is the shortest way from $a$ to $b$ while going through $X$. This makes sense because $X$ is compact, we need only show that $d$ is actually a metric. The only thing that isn’t clear is the triangle inequality, so here goes.

FACT 10: If $A \sse X$, with $A,X$ compact, and $f: A \rightarrow \R$ is a katetov function, then $\hat{f}: X \rightarrow \R$ is a katetov function that (obviously) extends $f$, where $$\hat{f}(x) := \inf_{a \in A} \{f(a)+d(a,x)\}$$

proof. If $a,b \in A$, the result is clear (although we didn’t really have to point out this case). Let $x,y \in X$.

$\|\hat{f}(x)-\hat{f}(y)\|$	$=\| \inf_{a \in A} \{f(a)+d(a,x)\} – \inf_{b \in A} \{f(b)+d(b,y)\}\|$
	$\leq \| \inf_{a \in A} \{f(a)+d(a,x)\} – (f(b_0)+d(b_0,y))\|$
	$\leq \| f(b_0)+d(b_0,x) – f(b_0)-d(b_0,y)\|$
	$\leq \|d(b_0,x) – d(b_0,y)\|$
	$\leq d(x,y)$
	$\leq d(x,a) + d(a,y) + f(a) + f(a), (\forall a \in A)$

Thus $d(x,y) \leq \inf_{a \in A} \{f(a)+d(a,x)\} + \inf_{a \in A} \{f(a)+d(a,y)\} = \hat{f}(x)+\hat{f}(y)$.

[Note that above, $b_0$ was chosen to be $\sup_{b\in A} \{f(b)+d(b,y)\}$.]

[QED]

Conclusion

There is the end of the easily digestible facts about the Urysohn space (that I know). It is interesting to me that I was able to get all of this information across to our seminar in a mere 90 minutes! As presented here the material seems so dense, but it really isn’t. I hope the pictures help you grok this stuff, because I really think that the ideas here are simple and the techniques are useful.

Most of this material was taken from the wonderful article “On the Geometry of Urysohn’s Universal Metric Space” by Julien Melleray (2007), and chapter 5 of “Dynamics of Infinite Dimensional Groups” by Vladimir Pestov (2006). Melleray’s paper contains very useful exercises, some of which I have included in this talk (as facts).

A partition theorem for finite trees

François G. Dorais — Sat, 03 Mar 2012 19:32:37 +0000

In a recent paper [1], Steven Gubkin, Daniel McDonald, Manuel Rivera, and I stumbled across what appears to be a new result in Ramsey theory. As the title of our paper suggests, this result was not exactly our main goal. Nevertheless, the result is very interesting so I feel it is a good idea to [...]

Indestructibility for Ramsey cardinals

Victoria Gitman — Thu, 01 Mar 2012 15:37:42 +0000

$
\newcommand{\p}{\mathbb{P}}
\newcommand{\q}{\mathbb{Q}}
\newcommand{\ORD}{{\rm ORD}}
\newcommand{\Add}{\text{Add}}
$
This is a talk at the Rutgers logic seminar, April 2, 2012.

Forcing is the main technique set theorists use for showing the consistency of various combinations of set theoretic properties. While a forcing extension $V[G]$ of a model $V$ of ${\rm ZFC}$ continues to satisfy ${\rm ZFC}$, it is not guaranteed that if $\kappa$ was a large cardinal in $V$, it will continue to be so in $V[G]$. For instance, forcing to collapse a large cardinal $\kappa$ to $\omega_1$, surely destroys the large cardinal. In order to establish the consistency of a large cardinal with a property obtainable by forcing, we need to argue that the large cardinal is indesructible by the forcing notion involved. This is precisely how we can determine something like the consistency strength of the ${\rm GCH}$ failing at a large cardinal. The standard toolkit of indestructibility techniques is designed for large cardinals that are characterized by the existence of elementary embeddings. This is true of measurable cardinals and most stronger large cardinals $\kappa$, which are characterized by the existence of elementary embeddings $j:V\to M$ from the universe $V$ into a transitive class $M$ with critical point $\kappa$ (the least ordinal moved) and whatever additional properties specific to the large cardinal. To argue, that, say, a measurable cardinal is not destroyed in a forcing extension $V[G]$ by $\p$, we work in $V[G]$ to extend $j$ to an elementary embedding $j:V[G]\to M[H]$. The target model must have the form $M[H]$, where $H$ is an $M$-generic filter for the poset $j(\p)$, by elementarity. The main theorem of the indestructibility toolkit, the lifting criterion, gives a prescription for extending $j$ to $V[G]$, so long as we can find an $M$-generic filter $H$ for $j(\p)$ containing $j“G$ as a subset. To obtain $H$, we use another crucial theorem from the toolkit, the diagonalization criterion, which generalizes the construction for obtaining generic filters for countable models to models of higher cardinality. Does this strategy apply to large cardinals weaker than a measurable cardinal?

Despite the fact that smaller large cardinals are most widely recognized for their combinatorial properties, many are characterized by the existence of elementary embeddings for “mini-universes" of set theory. The “mini-universes" are formally known as weak $\kappa$-models, $M\models {\rm ZFC}^-$ (${\rm ZFC}$ without powerset and with collection scheme instead of the replacement scheme) of size $\kappa$ and with $\kappa\in M$. Weakly compact cardinals have the simplest such characterization. A cardinal $\kappa$ is weakly compact if and only if every $A\subseteq\kappa$ is contained in a weak $\kappa$-model $M$ for which there exists an elementary embedding $j:M\to N$ with $N$ transitive and critical point $\kappa$. Ramsey cardinals are also in the league of smaller large cardinals having a little known but quite elegant elementary embeddings characterization. To motivate the characterization, let's recall that the existence of a definable elementary embedding $j:V\to M$ with critical point $\kappa$ is equivalent to the existence of a $\kappa$-complete ultrafilter on $\kappa$. Analogously, the existence of an elementary embedding $j:M\to N$ with critical point $\kappa$ is equivalent to the existence of an $M$-ultrafilter with a well-founded ultrapower. $M$-ultrafilters are very different creatures from the ultrafilters associated to measurable cardinals. For one thing they are, in almost all interesting cases, external to $M$. They measure just the subsets of $\kappa$ that are elements of $M$ and are $\kappa$-complete just for sequences in $M$. Since it is possible that a countable sequence of elements of an $M$-ultrafilter has an empty intersection, there is no reason to believe that the ultrapower by such an ultrafilter must be well-founded [1]. To attempt another broken analogy between $\kappa$-complete ultrafilters and $M$-ultrafilters, recall that the ultrapower construction with a $\kappa$-complete ultrafilter can be iterated $\ORD$-many times. The successor stages are constructed by taking the ultrapower by the image of the ultrafilter from the previous stage and direct limits are taken at limit stages. A famous theorem of Gaifman states that all such iterated ultrapowers are well-founded. To iterate the ultrapower construction by an $M$-ultrafilter $U$, we first need to decide what happens at successor stages since it is no longer possible to take the image of the externally existing ultrafilter. If $j:M\to N$ is the ultrapower map by $U$, we may try to obtain an $N$-ultrafilter $W$ by viewing $U$ as a predicate over $M$ and using Łós theorem:

$W=\{[f]_U\mid \{\alpha<\kappa\mid f(\alpha)\in U\}\}\in U$.

This clearly requires an additional assumption that sets of the form $\{\alpha<\kappa\mid f(\alpha)\in U\}$ be elements of $M$ and comes down to the property known as weak amenability that “$\kappa$-sized" pieces of $U$ are elements of $M$. Luckily if $U$ is weakly amenable, then so is $W$ and in fact, weak amenability propagates all along the iteration. So in possession of a weakly amenable $M$-ultrafilter, we may iterate the ultrapower construction. But the trouble does not stop there. For weakly amenable $M$-ultrafilters with a well-founded ultrapower, it does not follow that the iterates must all be well-founded. Indeed, we showed with Philip Welch that it is possible that an $M$-ultrafilter has anywhere from $1$ to any countable $\alpha$-many well-founded iterated ultrapowers (a part of the proof of Gaifman's theorem shows that if the first $\omega_1$-many iterated ultrapowers are well-founded than the rest must be well-founded as well) [2]. To ensure that all the iterates all well-founded, a sufficient condition due to Kunen is that the $M$-ultrafilter is countably complete, that is the intersection of any countable sequence of its elements is non-empty [3]. We can now fully appreciate the elementary embeddings characterization of Ramsey cardinals. A cardinal $\kappa$ is Ramsey if and only if every $A\subseteq\kappa$ is contained in a weak $\kappa$-model $M$ for which there exists a weakly amenable countably complete $M$-ultrafilter on $\kappa$. Pretty elegant right?

Unfortunately this elegant characterization does not submit itself easily to the indestructibility toolkit. The diagonalization criterion requires embeddings to be on $\kappa$-models, weak $\kappa$-models that are closed under sequences of length less than $\kappa$, which does not happen for the Ramsey embeddings. In fact, I showed that the existence of weakly amenable $M$-ultrafilters for $\kappa$-models pushes up the consistency strength beyond Ramsey cardinals. Also, it is not sufficient to simply extend the ultrapower embedding to the model $M[G]$. The extension is guaranteed by a simple argument to be the ultrapower by some $M[G]$-ultrafilter, but that ultrafilter is not guaranteed to be either weakly amenable or countably complete. While weak amenability usually follows easily, countable completeness presents a challenge.

In this talk, we will present a new diagonalization criterion for models without closure and give sufficient conditions for the $M[G]$-ultrafilter resulting from extending the ultrapower embedding to $M[G]$ to be countably complete. As a result, we will be able to show that Ramsey cardinals are indestructible by a variety of forcing notions.

Take a look at the slides here.

[1] H. Gaifman, “Elementary embeddings of models of set-theory and certain
subtheories.” Providence R.I.: Amer. Math. Soc., 1974, pp. 33-101.
[Bibtex]

@incollection {gaifman:ultrapowers,
AUTHOR = {Gaifman, Haim},
TITLE = {Elementary embeddings of models of set-theory and certain
subtheories},
BOOKTITLE = {Axiomatic set theory ({P}roc. {S}ympos. {P}ure {M}ath., {V}ol.
{XIII}, {P}art {II}, {U}niv. {C}alifornia, {L}os {A}ngeles,
{C}alif., 1967)},
PAGES = {33--101},
PUBLISHER = {Amer. Math. Soc.},
ADDRESS = {Providence R.I.},
YEAR = {1974},
MRCLASS = {02K15 (02H13)},
MRNUMBER = {0376347 (51 \#12523)},
MRREVIEWER = {L. Bukovsky},
}

[2] V. Gitman and P. D. Welch, “Ramsey-like cardinals II,” The Journal of Symbolic Logic, vol. 76, iss. 2, pp. 541-560, 2011.
[Bibtex]

@ARTICLE{gitman:welch,
AUTHOR= "Victoria Gitman and Philip D. Welch",
TITLE= "Ramsey-like cardinals {II}",
JOURNAL = {The Journal of Symbolic Logic},
VOLUME = {76},
YEAR = {2011},
NUMBER = {2},
PAGES = {541-560},
PDF={http://boolesrings.org/victoriagitman/files/2011/08/ramseylikecardinalsii.pdf},
EPRINT ={1104.4448},
ISSN = {0022-4812},
CODEN = {JSYLA6},
MRCLASS = {03E55},
MRNUMBER = {2830435 (2012e:03111)},
MRREVIEWER = {Bernhard A. K{\"o}nig},
DOI = {10.2178/jsl/1305810763},
URL = {http://dx.doi.org/10.2178/jsl/1305810763},
}

[3] K. Kunen, “Some applications of iterated ultrapowers in set theory,” Ann. Math. Logic, vol. 1, pp. 179-227, 1970.
[Bibtex]

@article {kunen:ultrapowers,
AUTHOR = {Kunen, Kenneth},
TITLE = {Some applications of iterated ultrapowers in set theory},
JOURNAL = {Ann. Math. Logic},
FJOURNAL = {Annals of Pure and Applied Logic},
VOLUME = {1},
YEAR = {1970},
PAGES = {179--227},
ISSN = {0168-0072},
MRCLASS = {02.68},
MRNUMBER = {0277346 (43 \#3080)},
MRREVIEWER = {A. M{\'a}t{\'e}},
}

Disqus

François G. Dorais — Thu, 01 Mar 2012 14:26:02 +0000

I have just enabled Disqus commenting system. It has lots of nice features, but there are a few bugs we need to fix — that’s part of the Boole’s Rings way… The most obvious issue is that MathJax currently doesn’t work in comments. I’m hoping that will get fixed very soon, so keep on using [...]

MA and its effect on Tree Partitions

Micheal Pawliuk — Tue, 28 Feb 2012 19:34:56 +0000

(This is the presentation I gave for Stevo Todorcevic’s course Combinatorial Set Theory on Feb 28, 2012. The material comes from Stevo’s 1983 paper “Partition Relations for Partially Ordered Sets”.)

In partition relations for ordinals, it has been established that:

Theorem (Erdos-Rado). $\omega_1 \rightarrow (\omega_1, \omega+1)^2$

Later it was shown that this is the best you can do, as the strengthenings are consistent:

Theorem(Hajnal). Under CH, $\omega_1 \not\rightarrow (\omega_1, \omega+2)^2$
Theorem (Todorcevic). Under PFA, for any countable ordinal $\alpha$, $\omega_1 \rightarrow (\omega_1, \alpha)^2$

Moving on, we can ask the same questions about non-special trees, which in some way are the tree analogue of “uncountable” or “large”.

Theorem (Todorcevic). Nonspecial Tree $\rightarrow$(Nonspecial Tree, $\omega+1)^2$

This is the analogue or the Erdos-Rado theorem.

Recall that a tree $T$ is nonspecial if $T \rightarrow (\omega)^1_\omega$, which means that any countable partition $T$ contains an infinite set. (This is a generalization of uncountable, because for countable sets you can always put one element per colour.)

We will show the following:

Theorem (Todorcevic). Under MA, for $T$ a tree with no uncountable chains and $\vert T \vert = 2^{\aleph_0}$ we have $T \not\rightarrow (T, \omega+2)^2$.

(Of course here, $T \not\rightarrow (T, \omega+2)^2$ should really be $T\not \rightarrow ($Nonspecial, $\omega)^2$. Throughout this talk I will assume that you understand that I am not demanding a copy of $T$ in the first colour.)

Through the proof we will actually get the superficially stronger result:

Theorem. Assume MA. Let $T$ be a tree with no uncountable chains and $\vert T \vert = 2^{\aleph_0}$. There is a partition $[T]^2 = K_0 \cup K_1$ such that

There is no nonspecial subtree $X\sse T$ such that $[X]^2 \sse K_0$;

There is no set $A \cup \{b,c\} \sse T$ such that $A < b<c$, otp $A=\omega$ and $A \times \{b,c\} \sse K_1$.

We will need the following standard results about MA and trees:

Lemma 1. Assume MA. Every Aronszajn tree $T$ of cardinality $\kappa<2^{\aleph_0}$ is special. (Thus nonspecial Aronszajn trees have cardinality $\geq 2^{\aleph_0}$)
Lemma 2. Every nonspecial tree $T$ of cardinality $2^{\aleph_0}$ can be pruned to a nonspecial tree $X$ such that for all $a \in X$, $X^a= \{t\in X : a \leq t\}$ is nonspecial.

To clarify, here an Aronszajn tree is a tree with uncountably many nodes, countable levels and no uncountable chains.

Recall. For $X$ a tree, $\hat{a}=\{t \in X : t \leq a\}$, the set of predecessors in $X$.

Lemma 3. Assume MA. Let $T$ be an nonspecial aronszajn tree (of cardinality $\kappa =2^{\aleph_0}$) with underlying set $\kappa$ that agrees with the tree ordering. (i.e. $\alpha <_T \beta$ implies $\alpha < \beta$). Fix $[\kappa]^\omega = \{A_\xi : \xi < \kappa\}$.
THEN there is a family $\{S_\alpha \sse \hat{\alpha}: \alpha < \kappa\}$ such that

$S_\alpha \cap S_\beta$ is finite for $\alpha \neq \beta$;

If $\xi < \alpha$, (and if $A_\xi \cap \hat{\alpha}$ is not covered by finitely many $S_\beta$ with $\beta<\alpha$) then $S_\alpha \cap A_\xi \neq \emptyset$.

Proofs.

[1]. This is a classical result of Baumgartner from 1970. The poset is what you might expect – finite approximation of a partition- but it is difficult to show this is ccc.

[2]. If there are a small (read: special-many) amount of nodes with special upwards cones, then we can simply remove them (as collection of special many special trees is special. Remember, special means small). The case where there are “nonspecial” many nodes with special upwards cones requires special attention which I will not go into here.

Helpful Diagram, no?

[3] This is proved by induction on $\alpha<\kappa$, and uses a very similar poset to the one to generate a MAD family with the property that all the elements of the mad family have large intersection with a prescribed family.

Now, on to the theorem!

proof. Let $T$ be an nonspecial aronszajn tree (of cardinality $\kappa =2^{\aleph_0}$) with underlying set $\kappa$ that agrees with the tree ordering. Assume by Lemma 2, that we have pruned $T$ properly. Fix $[\kappa]^\omega = \{A_\xi : \xi < \kappa\}$. Let $\{S_\alpha : \alpha < \kappa\}$ be a family as described in Lemma 3.

Define $[T]^2 = K_0 \cup K_1$ by $\{\beta, \alpha\} \in K_1$ iff $\beta \in S_\alpha$.

Claim 1. There are no 1-homogeneous subsets of order-type $\omega+2$.

Let $B = \{x_n : n<\omega\}\cup\{x_\omega, x_{\omega+1}\}$, with $[B]^2 \sse K_1$. Then $\vert S_{x_{\omega}}\cap S_{x_{\omega+1}} \vert = \vert \{x_n : n<\omega\} \vert = \aleph_0$, which violates the first condition of Lemma 3.

Claim 2. There is no 0-homogeneous nonspecial tree $X \sse T$.

Suppose not. By Lemma 1 and 2, we may assume that $X^a$ is nonspecial and of cardinality $\kappa$ for each $a \in X$.
For $\beta \in X$ define $$C_\beta := \{\alpha \in T :\vert S_\alpha \cap \hat{\beta} \cap X \vert = \aleph_0\}$$ Later these will be used to define a decomposition $X = \bigcup_{n<\omega} X_n$, where each $X_n$ is special, so $X$ will be special.

Fact 1. $C_\beta$ is finite for every $\beta \in X$.

Tree from Fact 1. Red means nonspecial. Grey means ordinal heights.

Suppose not. Let $C_\beta$ be infinite. In particular, $\hat{\beta}\cap X$ is infinite. Remember our enumeration of $[\kappa]^\omega$? Well, pick $\xi<\kappa$ such that $A_\xi = \hat{\beta}\cap X$.

Since we have MA, we get that $X^\beta$ is nonspecial and has cardinality $\kappa$. So find an $\alpha \in X^\beta$ such that $\xi < \alpha$.

Claim: $A_\xi \cap \alpha$ is not covered by finitely many $S_\gamma$ where $\gamma < \alpha$.

Recall that an infinite, almost disjoint family cannot finitely cover an infinite set. So $\{S_\gamma : \gamma < \alpha\}$ cannot finitely cover $A_\xi \cap \hat{\alpha} = A_\xi \cap \hat{\beta} = A_\xi$.

This yields a contradiction with property (ii) of the $S_\alpha$’s as there will be a $\gamma \in A_\xi \cap S_\alpha = \hat{\beta}\cap X\cap S_\alpha$ which means $\gamma \in S_\alpha$, ie $\{\gamma, \alpha\} \in K_1$, but we had assumed $[X]^2 \sse K_0$.

[QED, Fact 1]

So now we have a candidate for decomposing $X$ into countably many (hopefully small) pieces.

Let $X_n = \{\beta \in X: \vert C_\beta \vert = n\}$. The following fact will finish the proof.

Fact 2: Each $X_n$ is special.

Suppose that $X_n$ is nonspecial. By lemma 2, find a $\alpha \in X_n$ such that $T^{\alpha} \cap X_n$ is nonspecial. Let $C_\alpha = \{\beta_1, \dots, \beta_n\}$, written in increasing (ordinal) order.

The idea here is that if we extend to an element $\alpha^\prime \in T^{\alpha}\cap X_n$, then $C_\alpha = C_{\alpha^\prime}$; it stays fixed! We will extend twice.

Tree from Fact 2. Red means nonspecial. Grey means ordinal heights.

Since $T^{\alpha}\cap X_n$ is nonspecial, we may find an $\alpha^\prime \in T^{\alpha}\cap X_n$ such that $A := \{\gamma \in X_n : \beta_n < \gamma <_T \alpha^\prime\}$ is infinite. (There is a subtle point here about which ordering we are using.)

Remember that enumeration of $[\kappa]^\omega$? Let $\xi<\kappa$ be such that $A_\xi = A$.

Since $T^{\alpha^\prime} \cap X_n$ is nonspecial it has cardinality $\kappa$. Choose an $\alpha^{\prime\prime} < \kappa$ such that $\alpha^{\prime\prime} \in T^{\alpha^\prime} \cap X_n$.

Note $C_{\alpha^{\prime\prime}} = \{\beta_1, \dots, \beta_n\}$, so $A_\xi \cap S_\beta$ is finite for every $\beta<\alpha^{\prime\prime}$.

Thus by property (ii), $S_{\alpha^{\prime\prime}} \cap A_\xi \neq \emptyset$. i.e. There is a $\gamma \in X_n$ such that $\gamma < \alpha^{\prime\prime}$ and $\gamma \in S_{\alpha^{\prime\prime}}$, a contradiction.

[QED]

c.c.c. vs. the Knaster property

Assaf Rinot — Mon, 27 Feb 2012 05:01:24 +0000

After my previous post on Mekler’s characterization of c.c.c. notions of forcing, Sam, Mike and myself discussed the value of it . We noticed that a prevalent verification of the c.c.c. goes like this: given an uncountable set of conditions, apply the $\Delta$-system lemma, thin out the outcome some more (typically, a few iterations are needed) and then find two conditions in the thinned-out set which are already compatible. This successful method usually ends up showing more than the c.c.c. – it shows that the poset has the Knaster property (i.e., every uncountable set of conditions, contains an uncountable subset of pairwise compatible conditions). However, if the notion of forcing does not have the Knaster property, then this method is unlikely to do the job. A-ha! So, that’s perhaps the value of Mekler’s verification method – to establish c.c.c. for posets which are not better than that. We decided to test this conjecture.
Todorcevic was around during our conversation, so we asked him for an example of a poset which is c.c.c., but does not have the Kanster property (of course, a Souslin tree is such an example, but Souslin trees are c.c.c. by definition, and we wanted to examine the verification method). He provided the following hint.

Example. Suppose that $\mathfrak b=\omega_1$, and let $\overrightarrow f=\langle f_\alpha\mid\alpha<\omega_1\rangle$ witness that. That is:

$f_\alpha\in{}^\omega\omega$ is an increasing function for all $\alpha<\omega_1$;
$\{ n<\omega\mid f_\alpha(n)\ge f_\beta(n)\}$ is finite, whenever $\alpha<\beta<\omega_1$;
for every $g\in{}^\omega\omega$, there exists some $\alpha<\omega_1$ such that $\{ n<\omega\mid g(n)<f_\beta(n)\}$ is infinite, whenever $\alpha<\beta<\omega_1$.

Let $\alpha\unlhd\beta$ iff $\{ n<\omega\mid f_\alpha(n)>f_\beta(n)\}$ is empty. Now, let $\mathbb P$ be the collection of all finite antichains of the poset $\langle\omega_1,\unlhd\rangle$, ordered by inclusion.

So, I tried to use the Mekler method to verify that $\mathbb P$ is c.c.c., without much of success. (What about you? can you suggest such a proof?)
Eventually, I did find a proof, and this finding lead me to yet another characterization:

(Ridiculously trivial) Observtion. Suppose that $\mathbb P$ is a given notion of forcing. Then the following are equivalent:

$\mathbb P$ is c.c.c.;
for every large enough regular cardinal $\kappa$, there exists an elementary submodel $\mathcal N\prec(\mathcal H(\kappa),\in)$ with $\mathbb P\in\mathcal N$, that enjoys the following feature. For every uncountable $A\subseteq\mathbb P$ in $\mathcal N$, there exists $q\in A\setminus\mathcal N$, and $r\in A\cap\mathcal N$ which are compatible.

A second look at Mekler’s proof of the Devlin-Shelah theorem makes it clear that the latter characterization may be utilized to establish the c.c.c. of the Devlin-Shelah notion of forcing. So, we did not lose anything. Yet, the above (ridiculously trivial) characterization is the key to the proof that I came up with of the following.

Proposition. $\mathbb P$ is c.c.c.
Proof. Let $\mathcal N$ be an arbitrary elementary submodel of $(\mathcal H(\kappa),\in)$ for a large enough $\kappa$, such that $\overrightarrow f,\mathbb P\in\mathcal N$. Suppose that we are given an uncountable subset $A\subseteq\mathbb P$ in $\mathcal N$. Fix an arbitrary $q\in A\setminus\mathcal N$, and let us find some $r\in A\cap\mathcal N$ which is compatible with $q$.

Before we begin, we introduce some notation.

For all $\alpha<\beta$, denote $$\Delta(\alpha,\beta):=\min\{ m<\omega\mid m\le n<\omega\rightarrow(f_\alpha(n)\le f_\beta(n))\};$$
For all $\alpha<\beta$, denote $$\Gamma(\alpha,\beta):=\min\{ m<\omega\mid m\le n<\omega\rightarrow(f_\alpha(n)< f_\beta(n))\};$$
For all $p\in\mathbb P$, and $m<\omega$, let $h_p^m:|p|\rightarrow{}^m\omega$ be the function satisfying $h_p^m(i)=f_\alpha\restriction m$ for the unique $\alpha\in p$ such that $|p\cap\alpha|=i$.

Here we go. Let $\rho=q\cap\mathcal N$. Then $\rho$ is a finite subset of $\mathcal N$, and hence in $\mathcal N$. Let $$S:=\{ \nu<\omega_1\mid \exists p\in A(p\cap\nu=\rho)\}.$$ Since $S\in\mathcal N$, and $\mathcal N\cap\omega_1\in S$ (as witnessed by $q$), we infer that $S$ is stationary. In particular, we may fix some $\nu\in S$ above $\max(q)$. Pick $p\in A\setminus\mathcal N$ such that $p\cap\nu=\rho$. Since $p\not\in\mathcal N$, we get that $p\neq\rho$ and $p\neq q$, so let $$m:=\max\{\Gamma(\alpha,\gamma)\mid \alpha<\gamma, \alpha\in q, \gamma\in p\}.$$

Put $h:=h^{m+1}_p$. Then $h\in\mathcal N$, and $p$ is a member of the following set (which lies in $\mathcal N$, as well): $$A(h,\rho):=\{ r\in A\mid h^{m+1}_r=h\ \& \exists\alpha<\omega_1(r\cap\alpha=\rho)\}.$$ In particular, we may pick some $r\in A(h,\rho)\cap\mathcal N$.
As $h^{m+1}_r=h^{m+1}_p$, we get that $|r|=|p|$, so let $\pi:r\rightarrow p$ be the order-preserving isomorphism. Then $\pi\restriction\rho$ is the identity map, and $f_\beta\restriction(m+1)=f_{\pi(\beta)}\restriction(m+1)$ for all $\beta\in r$.

We claim that $r$ and $q$ are compatible. Namely, that $\Delta(\beta,\alpha)>0$ for all $\beta<\alpha$ in $r\cup q$. Suppose that $\beta<\alpha$ are in $r\cup q$. For $\beta,\alpha\in r$ or $\beta,\alpha\in q$, it is immediate that $\Delta(\beta,\alpha)>0$, so suppose that $\beta\in r\setminus q$ and $\alpha\in q\setminus r$. As $r\cap q=\rho$, we get that $\beta<\alpha$. Let $\gamma:=\pi(\beta)$. By $f_\beta\restriction(m+1)=f_\gamma\restriction(m+1)$ and $\Gamma(\alpha,\gamma)\le m$, we get that $f_\alpha(m)< f_\gamma(m)=f_\beta(m)$, and hence $\Delta(\beta,\alpha)>m\ge 0$. $\square$

Note that essentially the same proof as above shows that $\mathbb P$ is productively c.c.c., and also that this works to any uncountable sequence of reals $\overrightarrow f$. So, what’s the role of $\mathfrak b=\aleph_1$? It is here to insure the following.

Proposition. $\mathbb P$ does not have the Knaster property.
Proof. Suppose that $A$ is an uncountable subset of $\mathbb P$. We shall find two conditions in $A$ which are incompatible. For this, we may already assume that $A$ forms a $\Delta$-system with a root $\rho$. In particular, $p\mapsto\min(p\setminus\rho)$ is an injection, and $B=\{ \min(p\setminus\rho)\mid p\in A\}$ is an uncountable subset of $\omega_1$. If we can find $\alpha<\beta$ in $B$ for which $\Delta(\alpha,\beta)=0$, then for the corresponding conditions $p,q\in A$ (i.e., $\alpha=\min(p\setminus\rho), \beta=\min(q\setminus\rho))$, we get that $p\cup q$ is not an antichain in the poset $\langle \omega_1,\unlhd\rangle$, and hence, $p$ and $q$ are incompatible.

Here we go. Fix a large enough regular cardinal $\kappa$, and then a countable elementary submodel $\mathcal N\preceq(H(\kappa),\in)$ with $B,\overrightarrow f\in\mathcal N$. Let $\delta:=\min(B\setminus\mathcal N)$. Evidently, $B\setminus(\delta+1)=\bigcup_{n<\omega}\{\beta\in B\setminus(\delta+1)\mid \Delta(\delta,\beta)=n\}$, so let us fix some $n<\omega$ such that $B_n:=\{\beta\in B\setminus(\delta+1)\mid \Delta(\delta,\beta)=n\}$ is uncountable. By the choice of $\overrightarrow f$ (more specifically, by item (3) there), we get that the set $$M:=\{ m<\omega\mid \sup\{ f_\beta(m)\mid \beta\in B_n\}=\omega\}$$ is non-empty (in fact, infinite), so consider its minimal element, $m:=\min(M)$.
For $t\in{}^m\omega$, denote $B^t_n:=\{ \beta\in B_n\mid t\subseteq f_\beta\}$. By minimality of $m$, the set $\{ t\in{}^m\omega\mid B^t_n\neq\emptyset\}$ is finite, so we can easily find some $t\in{}^m\omega$ such that $$\sup\{ f_\beta(m)\mid \beta\in B^t_n\}=\omega.$$ Since $\{ \beta\in B\mid t\subseteq f_\beta\}$ is a non-empty set that lies in $\mathcal N$, let us fix some $\alpha\in\mathcal N\cap B$ such that $t\subseteq f_\alpha$. Put $k:=\Delta(\alpha,\delta)$, and then pick $\beta\in B^t_n$ such that $f_\beta(m)>f_\alpha(k+n)$. Of course, $\alpha<\delta<\beta$.
We claim that $\Delta(\alpha,\beta)=0$. This is best seen by dividing into three cases:

if $i<m$, then $f_\alpha(i)=t(i)=f_\beta(i)$;
if $m\le i\le k+n$, then $f_\alpha(i)\le f_\alpha(k+n)<f_\beta(m)\le f_\beta(i)$ (recall that the elements of $\overrightarrow f$ are increasing functions!);
if $k+n<i<\omega$, then $\Delta(\alpha,\delta)=k<i$ and $f_\alpha(i)\le f_\delta(i)$, as well as $\Delta(\delta,\beta)=n<i$ and $f_\delta(i)\le f_\beta(i)$. Altogether, $f_\alpha(i)\le f_\beta(i)$. $\square$

We conclude this post with an awkward proof of the following.

Fact. $\text{MA}_{\aleph_1}$ entails $\mathfrak b>\omega_1$.
Proof. Suppose not. Then $\text{MA}_{\aleph_1}$ is consistent together with the existence of a poset $\mathbb P=\langle P,\preceq\rangle$ of size $\aleph_1$ which is productively $c.c.c.$, but not Knaster. Define a poset $\mathbb Q=\langle Q,\le\rangle$, by $Q:={}^{<\omega}P$, and $q_1\le q_2$ (i.e., $q_2$ extends $q_1$) iff $\text{dom}(q_1)\subseteq \text{dom}(q_2)$, and $q_1(i)\preceq q_2(i)$ for all $i\in\text{dom}(q_1)$. Since $\mathbb P$ is productively $c.c.c$, $\mathbb Q$ is $c.c.c.$. Also, it a trivial fact that for any $p\in P$, and $q\in Q$, $q{}^\frown p\in Q$. So, $D_p:=\{ q\in Q\mid p\in\text{rng}(q)\}$ is dense in $\mathbb Q$, for all $p\in P$.
Finally, by $\text{MA}_{\aleph_1}$, let $G\subseteq Q$ be a directed set that meets any of the sets from the sequence $\langle D_p\mid p\in \mathbb P\rangle$. Denote $P_i:=\{ q(i)\mid q\in G\}$ for all $i<\omega$. Then $P=\bigcup_{i<\omega}P_i$. Since $G$ is $\le$-directed, $P_i$ is $\preceq$-directed for all $i<\omega$. It follows that for every uncountable $A\subseteq P$, there exists some $i<\omega$ such that $A\cap P_i$ is an uncountable subset of $A$ whose elements are pairwise compatible. That is, $\mathbb P$ has the Knaster property, contradicting the very choice of $\mathbb P$. $\square$

Generalized separation principles

François G. Dorais — Sun, 26 Feb 2012 16:09:16 +0000

$\newcommand{\RCA}{\mathsf{RCA}_0}\newcommand{\WKL}{\mathsf{WKL}_0}\newcommand{\ACA}{\mathsf{ACA}_0}\newcommand{\DNR}{\mathsf{DNR}}\newcommand{\dom}{\mathrm{dom}}\newcommand{\Sep}{\mathsf{S}}$ It is well-known that computability theory and reverse mathematics have very strong ties. Indeed, the base theory $\RCA$ used in reverse mathematics was designed as the minimal theory that can adequately formulate and prove the basic results of computability theory. Based on the Church–Turing thesis, this is a very reasonable way to capture the [...]

Special uncountable trees

Samuel Coskey — Fri, 24 Feb 2012 04:49:49 +0000

In a recent post, Assaf asked whether Mekler’s characterization of ccc posets can be used to give an alternative proof of Baumgartner’s theorem that the natural poset to specialize an Aronszajn tree is ccc. In this post, I’ll recall what that poset is, and Baumgartner’s famous argument.

Aronszajn trees

A classic observation in finite combinatorics, known as König’s lemma, states that if a tree has the following properties

every level is finite, and
every chain is finite

then the whole tree must be finite. Surprisingly, if we allow ourselves to look at set-theoretic trees (whose levels are indexed by ordinals instead of natural numbers) then even the simplest generalization of this statement fails badly. Namely, there exist trees with the properties

every level is countable, and
every chain is countable, and yet
the tree is uncountable

Such a tree is called an Aronszajn tree. They can be built using a simple recursive construction.

Special Aronszajn trees

Given the motivation from König’s lemma, it is just as natural to ask whether there exists an uncountable tree with the stronger properties:

every antichain is countable, and
every chain is countable

Such trees are called Suslin trees and they are of historical interest thanks to a connection with topology. (Specifically, the non-existence of Suslin trees is equivalent to the statement that $\mathbb R$ is the unqiue complete, dense, linear order with the topological ccc.)

At first glance, it is not easy to tell what is the relationship between Suslin trees and Aronszajn trees. Of course, every Suslin tree is Aronszajn. But it turns out that there is a key difference. Notice that by definition, a Suslin tree cannot be written as the union of countably many antichains. But if you carry out the recursive construction indicated above, the Aronszajn tree you get can always be written as the union of countably many antichains. Trees with this property are called special.

The poset to specialize an Aronszajn tree

Let us fix an Aronszajn tree $T$ and consider the natural poset of finite approximations to a specializing function for $T$. More precisely, $p\in\mathrm P$ iff:

$p$ is a finite partial function $T\rightarrow\omega$
for all $n$, $p^{-1}(n)$ is an antichain

The partial ordering on $\mathrm P$ is simply inclusion.

Without getting into the details of forcing, let it suffice to say that after forcing with $\mathrm P$, the tree $T$ will be special in the extension. Let us also note that it is highly desirable to verify that $\mathrm P$ is ccc. This implies that forcing with $\mathrm P$ does not change cardinal arithmetic, and also that $\mathrm P$ can be iterated. So for instance it follows from this that there is a model (such as a model of MA) in whith every Aronszajn tree is special. Thus, in this model there are no Suslin trees!

Baumgartner’s argument

In this last section, we assume the reader is familiar with the $\Delta$-system lemma, outlined in Mike’s post here, a standard tool for showing posets are ccc.

We now begin the proof that $\mathbb P$ has the ccc. Suppose towards a contradiction that $\mathbb P$ has an uncountable antichain $A$. This simply means that for distinct $p,q\in A$, some fiber of $p\cup q$ is not an antichain in $T$: there exist $t,u\in T$ such that $p(t)=q(u)$ and $t,u$ are compatible in $T$.

This gives us quite a lot of compatible elements of $T$—what we want to do is organize them until we can find an uncountable chain, which would be a contradiction because $T$ is an Aornszajn tree. So a little bit of combinatorics will be needed; in this case an ultrafilter on $\omega_1$.

First, however, let us suppose without loss of generality that the elements of $A$ all have the same size. Furthermore, using the $\Delta$-system lemma in the standard way, we can suppose that the domains of the elements of $A$ are pairwise disjoint.

Now comes the most desperate part of the argument. For each $p\in A$ we need names for elements in the domain of $p$, so write $\dom(p)=\set{t^p_1,\ldots,t^p_N}$. What we know is that for each distinct pair $p,q\in A$ there exist $k,l$ such that $p(t^p_k)=q(t^q_l)$ and $t^p_k,t^q_l$ are compatible in $T$. It seems natural to try to stabilize $k$ and $l$, but since these values depend on both $p$ and $q$, this is not possible in general. The best we can achieve under these circumstances in the following.

Now comes the most clever part of the argument. Let $\mathcal U$ be a (nonprincipal) ultrafilter on $A$. Then for each $p$ there exist $k,l$ such that the following set lies in $\mathcal U$.
\[A_{p,k,l}=\set{q\mid p(t^p_k)=q(t^q_l)\text{ and $t^p_k,t^q_l$ are compatible in $T$}}
\]Without loss of generality, we can suppose that these $k,l$ are the same for each $p$.

To achieve a contradiction, we will now show that the elements $t^p_k$ form a chain in $T$. Indeed, given $t^p_k$ and $t^q_k$, there are uncountably many elements $u\in A_{p,k,l}\cap A_{q,k,l}$, and these have the property that $t^u_l$ is compatible with both $t^p_k$ and $t^q_k$. Hence there is some such $t^u_l$ above both $t^p_k$ and $t^q_k$. Since $T$ is a tree, it follows that $t^p_k,t^q_k$ are compatible in $T$. This concludes the proof!

Dushnik-Miller for regular cardinals (part 3)

Assaf Rinot — Thu, 23 Feb 2012 04:14:04 +0000

Here is what we already know about the Dushnik-Miller theorem in the case of $\omega_1$ (given our earlier posts on the subject):

$\omega_1\rightarrow(\omega_1,\omega+1)^2$ holds in ZFC;
$\omega_1\rightarrow(\omega_1,\omega+2)^2$ may consistently fail;
$\omega_1\rightarrow(\omega_1,\omega_1)^2$ fails in ZFC.

In this post, we shall provide a proof of Todorcevic’s theorem that PFA implies $\omega_1\rightarrow(\omega_1,\alpha)^2$ for all $\alpha<\omega_1$.

We commence with a sequence of lemmas.

Lemma 1. Suppose that $\langle I_\beta\mid \beta<\omega_1\rangle$ is a given sequence of sets, with $I_\beta\subseteq\beta$ for all $\beta<\omega_1$. For $Z\subseteq\omega_1$, denote $$\mathcal I(Z):=\{ X\in[Z]^{\le\aleph_0}\mid |\{\beta\in Z\mid X\cap I_\beta\text{ is infinite}\}|\le\aleph_0\}.$$

If $A\subseteq Z\subseteq\omega_1$ are uncountable sets, and $[A]^{\aleph_0}\subseteq\mathcal I(Z)$, then there exists some uncountable $B\subseteq A$ such that $\alpha\not\in I_{\beta}$ for all $\alpha<\beta$ in $B$.
Proof. As $A\cap\delta\in\mathcal I(Z)$ for all $\delta<\omega_1$, we may recurisvely construct a strictly-increasing function $f:\omega_1\rightarrow A$ such that $f[\beta]\cap I_{f(\beta)}$ is finite for all $\beta<\omega_1$. Let $A_1:=f“[\omega_1]$. Then $A_1$ is an uncountable subset of $A$, and $x_\beta:=A_1\cap I_\beta$ is finite for all $\beta\in A_1$. There are two cases to consider:

If $\{ x_\beta\mid \beta\in A_1\}$ is countable, then $B:=A_1\setminus\bigcup\{ x_\beta\mid \beta\in A_1\}$ is an uncountable subset with the desired property;
Otherwise, let $A_2$ be an uncountable subset of $A_1$ for which $\{ x_\beta\mid \beta\in A_2\}$ forms a $\Delta$-system with root $r$. As the sets in $\{ (x_\beta\setminus r)\mid \beta\in A_2\}$ are mutually disjoint, we get that $\sup\{ \min(x_\beta\setminus r)\mid \beta\in A_2\}=\omega_1$, so it is possible to recursively construct an uncountable subset $B\subseteq(A_2\setminus r)$ such that $\alpha\not\in x_\beta$ for all $\alpha<\beta$ in $B$. $\square$

Lemma 2. If $\mathfrak b>\aleph_1$, then $\mathcal I(Z)$ is a P-ideal for every $Z\subseteq\omega_1$.

[Reminder. The cardinal $\mathfrak b$ is defined as the smallest size of a family $\mathcal B\subseteq{}^\omega\omega$ with the property that for every $f:\omega\rightarrow\omega$, there exists some $b\in\mathcal B$ such that $\{ n<\omega\mid f(n)<b(n)\}$ is infinite. We say that an ideal $\mathcal I$ is a P-ideal if for every $\mathcal A\in[\mathcal I]^{\aleph_0}$, there exists some $B\in\mathcal I$ such that $B\setminus A$ is finite for all $A\in\mathcal A$. (slang: every countable collection of $\mathcal I$-sets, admits a pseudounion within $\mathcal I$).]

Proof of Lemma 2. If $Z$ is countable, then $\mathcal I(Z)=\mathcal P(Z)$, and we are done. Now, suppose that $Z$ is uncountable, and that we are given a countable subcollection $\{ X_n\mid n<\omega\}\subseteq\mathcal I$, and let us show that it admits a pseudounion. As $\mathcal I$ is downard closed, we shall avoid trivialities and assume that $\{ X_n\mid n<\omega\}$ are mutually disjoint. Let $\delta<\omega_1$ be large enough so that $$\bigcup_{n<\omega}\{\beta<\omega_1\mid X_n\cap I_\beta\text{ is infinite}\}\subseteq\delta.$$ Fix a bijection $\psi:\omega\rightarrow \bigcup_{n<\omega}X_n$ . Then for all $\beta\in Z\setminus\delta$, define $f_\beta:\omega\rightarrow\omega$ by letting for all $n<\omega$: $$f_\beta(n):=\min\{ m<\omega\mid X_n\cap I_\beta\subseteq \psi“m\}.$$

As $\mathfrak b>\omega_1$, we may pick a function $f:\omega\rightarrow\omega$ such that $\{ n<\omega\mid f_\beta(n)> f(n)\}$ is finite, for all $\beta\in Z\setminus\delta$. Now, let $X:=\bigcup\{X_n\setminus \psi[f(n)]\mid n<\omega\}$. Then, for every $n<\omega$, $X_n\setminus X\subseteq \psi[f(n)]$ is finite. Assume towards a contradiction that $X\not\in\mathcal I$. Then, in particular, there exists some $\beta\in Z\setminus\delta$ such that $I_\beta\cap X$ is infinite. As $f_\beta\le^* f$ while $I_\beta\cap X_n$ is finite for all $n<\omega$, let us pick a large enough $n<\omega$ such that $f_\beta(n)<f(n)$ and $(I_\beta\cap X)\cap X_n\neq\emptyset$. Pick $\gamma\in I_\beta\cap X\cap X_n$, and let $m:=\psi^{-1}(\gamma)$. Since $\gamma\in I_\beta\cap X_n$, we get that $f_\beta(n)>m$. In particular, $f(n)>m$, and so by definition of $X$, we get that $\psi(m)\not\in X$, contradicting the choice of $\gamma=\psi(m)$ in $X$. $\square$

Lemma 3. Suppose that $A\subseteq Z\subseteq\omega_1$ satisfies $[A]^{\aleph_0}\cap\mathcal I(Z)=\emptyset$. Denote $$\mathcal F(A,Z):=\left\{ F\in[Z]^{<\omega}\mid \left(A\setminus \bigcup_{\beta\in F}I_\beta\right)\text{ is finite }\right\}.$$ If $\mathfrak p>\aleph_1=|Z|$, then $\sup\{\min(F)\mid F\in\mathcal F(A,Z)\}=\omega_1$.

[Reminder. The cardinal $\mathfrak p$ is defined as the smallest size of a family $\mathcal A$ of subsets of $\omega$ such that $|\bigcap\mathcal F|=\aleph_0$ for all finite $\mathcal F\subseteq\mathcal A$, while $\mathcal A$ does not admit a pseudointersection. (A pseudointersection for $\mathcal A$ is an infinite set $B$ such that $B\setminus A$ for all $A\in\mathcal A$). Note that $\mathfrak b\ge\mathfrak p$.]

Proof of Lemma 3. Suppose not, and let $\delta:=\sup\{\min(F)\mid F\in\mathcal F(A,Z)\}$. Then, the intersection of any finite family of sets from $\{ A\setminus I_\beta\mid \beta\in Z\setminus\delta+1\}$ is infinite. Then, by $\mathfrak p>\omega_1$, there must exist a pseudointersection $B\subseteq A$, namely, $B\setminus( A\setminus I_\beta)$ is finite whenever $\beta\in Z\setminus\delta+1$. In particular, $\{ \beta\in Z\mid B\cap I_\beta\text{ is finite}\}$ is co-countable, and then $B\in\mathcal I(Z)$, contradicting the fact that $B\subseteq A$, and $[A]^{\aleph_0}\cap\mathcal I(Z)=\emptyset$. $\square$

Lemma 4 (Laver, 1973). For every nonzero $\alpha<\omega_1$, and every martix $\langle A_{i,\zeta}\mid i<n,\zeta<\kappa\rangle$, if all of the following conditions are met:

$n<\omega$;
$\kappa<\mathfrak p$;
$\bigcup_{i<n}A_{i,\zeta}=\omega^\alpha$ for all $\zeta<\kappa$,

then, there exists a function $f:\kappa\rightarrow n$ and a set $B\subseteq\omega^\alpha$ of order-type $\omega^\alpha$ such that $B\subseteq^* A_{f(\zeta),\zeta}$ for all $\zeta<\kappa$. (By $B\subseteq^*A$, we mean that $\sup(B\setminus A)<\sup(B)$.)

Proof. By induction on $\alpha$. For the base case $\alpha=1$, fix a uniform ultrafilter $\mathcal F$ over $\omega$. Then, let $f:\kappa\rightarrow n$ be a function such that $A_{f(\zeta),\zeta}\in\mathcal F$ for all $\zeta<\kappa$. As $\kappa<\mathfrak p$, the collection $\{ A_{f(\zeta),\zeta}\mid \zeta<\kappa\}$ must admit a pseudointersection.
Next, suppose that $1<\alpha<\omega_1$, and that the lemma holds for all $\beta<\alpha$. Let $\langle \beta_m\mid m<\omega\rangle$ be a non-decreasing sequence of ordinals such that $1\le\beta_m<\alpha$ for all $m<\omega$, and $\sum_{m<\omega}\omega^{\beta_m}=\omega^\alpha$. Denote $$A^m_{i,\zeta}:=A_{i,\zeta}\cap\left(\left(\sum_{k\le m}\omega^{\beta_k}\right)\setminus\left(\sum_{k< m}\omega^{\beta_k}\right)\right).$$ Then $A^m_{i,\zeta}$ is simply the $m_{th}$-section of $A_{i,\zeta}$. In particular:

$A_{i,\zeta}=\uplus_{m<\omega}A^m_{i,\zeta}$ for all $i<n$ and $\zeta<\kappa$;
$\text{otp}(\bigcup_{i<n}A^m_{i,\zeta})=\omega^{\beta_m}$ for all $m<\omega$, and $\zeta<\kappa$.

By the induction hypothesis, for every $m<\omega$, there exists a function $f_m:\kappa\rightarrow n$ such that $\left\{ A^m_{f_m(\zeta),\zeta}\mid \zeta<\kappa\right\}$ admits a pseduointersection $B_m$ of order-type $\omega^{\beta_m}$. In particular, $\sup(B_m)=\sum_{k\le m}\omega^{\beta_k}$. For all $m<\omega$, let $\langle \beta^j_m\mid j<\omega\rangle$ be an increasing sequence of ordinals that converges to $\sum_{k\le m}\omega^{\beta_k}$. Then, for all $\zeta<\kappa$, define a function $h_\zeta:\omega\rightarrow\omega$, by letting:

$$h_\zeta(m):=\min\left\{ j<\omega\mid \left(B_m\setminus A^m_{f_m(\zeta),\zeta}\right)\subseteq \beta^j_m\right\},\quad(m<\omega).$$

As $\kappa<\mathfrak p\le\mathfrak b$, let us pick function $h:\omega\rightarrow\omega$ such that $\{ m<\omega\mid h(m)<h_\zeta(m)\}$ is finite for all $\zeta<\kappa$. Next, we shall utilize the fact that $\{ f_m(\zeta)\mid m<\omega\}$ is finite for all $\zeta<\kappa$. Fix a uniform ultrafilter $\mathcal F$ over $\omega$, and then a function $g:\kappa\rightarrow n$ such that for all $\zeta<\kappa$:$$\{ m<\omega \mid f_m(\zeta)=g(\zeta)\}\in\mathcal F.$$
As $\kappa<\mathfrak p$, we may then find an infinite $M\subseteq\omega$ such that $\{ m\in M\mid f_m(\zeta)\neq g(\zeta)\}$ is finite for all $\zeta<\kappa$. Finally, appeal to $h$, and $M$, by letting $$B:=\biguplus\{ B_m\setminus \beta_m^{h(m)}\mid m\in M\}.$$ Then:

$\text{otp}(B_m\setminus \beta_m^{h(m)})=\text{otp}(B_m)=\omega^{\beta_m}$ for all $m\in M$, since $\beta_m^{h(m)}<\sum_{k\le m}\omega^{\beta_k}=\sup(B_m)$ ;
$\text{otp}(B)=\sum_{m\in M}\omega^{\beta_m}=\omega^\alpha$, since $\sup(M)=\omega$.

Thus, we are left with verifying that $B$ is indeed a pseudointersection of $\{ A_{g(\zeta),\zeta}\mid \zeta<\kappa\}$. For this, fix an arbitrary $\zeta<\kappa$. Let $t<\omega$ be large enough so that $$t>\max\{ m\in M\mid h(m)<h_\zeta(m)\ \&\ f_m(\zeta)\neq g(\zeta)\}.$$ We claim that $(B\setminus A_{g(\zeta),\zeta})\subseteq\sum_{k\le t}\omega^{\beta_k}<\omega^\alpha$. Suppose not, and fix $\delta\in (B\setminus A_{g(\zeta),\zeta})$ above $\sum_{k\le t}\omega^{\beta_k}$. Let $m\in M$ be the unique integer such that $\delta\in B_m \setminus \beta_m^{h(m)}$. Then $m>t$ and hence $h_\zeta(m)\le h(m)$, and $f_m(\zeta)=g(m)$. It follows that $$\delta\in\left(B_m\setminus A_{g(\zeta),\zeta}\right)\subseteq\left(B_m\setminus A^m_{g(\zeta),\zeta}\right)=\left(B_m\setminus A^m_{f_m(\zeta),\zeta}\right)\subseteq \beta^{h_\zeta(m)}_m\subseteq \beta^{h(m)}_m,$$contradicting the fact that $\delta\in B$. $\square$

Definition (Todorcevic). The P-ideal Dichotomy (abbrevated PID) asserts that for every uncountable set $Z$, and every P-Ideal $\mathcal I$ over $[Z]^{\le\aleph_0}$, exactly one of the following holds:

There exists an uncountable $A\subseteq Z$ such that $[A]^{\aleph_0}\subseteq\mathcal I$;
There exists a sequence $\langle Z_n\mid n<\omega\rangle$ such that $\bigcup_{n<\omega}Z_n=Z$, and $[Z_n]^{\aleph_0}\cap\mathcal I=\emptyset$ for all $n<\omega$.

Now, we are ready to prove the main result of this post.

Theorem (Todorcevic, 1983). Assume PID+($\mathfrak p>\omega_1$). Then $\omega_1\rightarrow(\omega_1,\alpha)^2$ holds for all $\alpha<\omega_1$.
Proof. We prove $\omega_1\rightarrow(\omega_1,\omega^\alpha)^2$ by induction on $\alpha<\omega_1$. The base case $\alpha=1$, is covered by the original Dushnik-Miller theorem. Next, suppose that $1<\alpha<\omega_1$, and that $\omega_1\rightarrow(\omega_1,\omega^\beta)^2$ holds for all $\beta<\alpha$, and let us establish $\omega_1\rightarrow(\omega_1,\omega^\alpha)^2$. For this, suppose that $c:[\omega_1]^2\rightarrow 2$ is a given coloring such that $c“[B]^2\neq\{0\}$ for every uncountable subset $B\subseteq\omega_1$.
For all $\beta<\omega_1$, denote $I_\beta:=\{\gamma<\beta\mid c(\gamma,\beta)=1\}$. Then $c(\gamma,\beta)=0$ iff $\gamma\not\in I_\beta$. In particular, we infer from Lemma 1 that for every uncountable $A\subseteq Z\subseteq\omega_1$, $[A]^{\aleph_0}\nsubseteq\mathcal I(Z)$. Namely, the first alternative of the P-ideal dichotomy fails for $\mathcal I(Z)$. As we assume PID$+(\mathfrak p>\omega_1)$, and since $\mathfrak b\ge\mathfrak p$, we infer from Lemma 2 that for every uncountable $Z\subseteq\omega_1$, there exists a sequence $\langle Z_n\mid n<\omega\rangle$ such that $\bigcup_{n<\omega}Z_n=Z$, and $[Z_n]^{\aleph_0}\cap\mathcal I(Z)=\emptyset$ for all $n<\omega$.
Altogether, we conclude that for every uncoutnable set $ Z\subseteq\omega_1$, there exists an uncountable subset $Y\subseteq Z$ such that $[Y]^{\aleph_0}\cap\mathcal I(Z)=\emptyset$.

Next, fix a non-decreasing sequence of ordinals $\langle \alpha_m\mid m<\omega\rangle$ such that $1<\alpha_m<\alpha$ for all $m<\omega$ and $\sum_{m<\omega}\omega^{\alpha_m}=\omega^\alpha$. We shall construct by recursion on $m<\omega$, a sequence $\langle (A_m,B_{m},\tau_{m},Z_m,Y_m,X_m)\mid m<\omega\rangle$, such that all of the following holds for all $m<\omega$;

$B_{m}\subseteq A_m\subseteq Y_m\subseteq Z_m\supseteq X_m\supseteq Z_{m+1}$;
$\text{otp}(B_m)=\text{otp}(A_m)=\omega^{\alpha_m}$;
$\text{otp}(Y_m)=\text{otp}(Z_m)=\text{otp}(X_m)=\omega_1$;
$\tau_{m}<\sup(B_{m})<\min(B_{m+1})$;
$c“[B_{m}]^2=\{1\}$;
$c“[(B_{m}\setminus\tau_{m})\times X_{m}]=\{1\}$.

Let $Z_0=\omega_1$. Then, let $Y_0$ be an uncountable subset of $Z_0$ such that $[Y_0]^{\aleph_0}\cap\mathcal I(Z_0)=\emptyset$. By the induction hypothesis, pick $A_0\subseteq Y_0$ of order-type $\omega^{\alpha_0}$ such that $c“[A_0]^2=\{1\}$. Since $[A_0]^{\aleph_0}\cap\mathcal I(Z_0)=\emptyset$, we get from Lemma 3 that $\mathcal F(A_0,Z_0)$ contains an uncountable family $\mathcal F_0\subseteq[Z_0]^{<\omega}$ consisting of mutually-disjoint sets of some fixed size $n$. Then, by Lemma 4, there exists a choice function $f_0:\mathcal F_0\rightarrow Z_0$ and $B_0\subseteq A_0$ of order-type $\omega^{\alpha_0}$ such that $B_0\subseteq^* I_{f_0(F)}$ for all $F\in\mathcal F_0$. Fix $\tau_0<\sup(B_0)$, and an uncountable subset $X_0\subseteq\text{rng}(f_0)$ such that $(B_0\setminus\tau_0)\subseteq I_\beta$ for all $\beta\in X_0$.
Now, suppose that $m<\omega$ is given, and $(A_m,B_{m},\tau_{m},Z_m,Y_m,X_m)$ has already been defined. Let us define $(A_{m+1},B_{m+1},\tau_{m+1},Z_{m+1},Y_{m+1},X_{m+1})$. Put $Z_{m+1}:=X_m$. Pick $Y_{m+1}\in[Z_{m+1}\setminus(\sup(B_m)+1)]^{\aleph_1}$ such that $[Y_{m+1}]^{\aleph_0}\cap\mathcal I(Z_{m+1})=\emptyset$. Pick $A_{m+1}\subseteq Y_{m+1}$ of order-type $\omega^{\alpha_{m+1}}$ such that $c“[A_{m+1}]^2=\{1\}$. Then, use Lemma 4 to find $B_{m+1}\subseteq A_{m+1}$ of order-type $\omega^{\alpha_{m+1}}$, $\tau_{m+1}<\sup(B_{m+1})$, and an uncountable $X_{m+1}\subseteq Z_{m+1}$ such that $c“[(B_{m+1}\setminus\tau_{m+1})\times X_{m+1}]^2=\{1\}$. This completes the construction.

Finally, let $B:=\bigcup_{n<\omega}(B_n\setminus\tau_n)$. Then $\text{otp}(B)=\omega^\alpha$, and $c“[B]^2=\{1\}$. $\square$

Subposets of small dimension

François G. Dorais — Sat, 18 Feb 2012 16:18:28 +0000

Behind the scenes at Boole’s Rings, there was some talk on asking more open questions related to our research. I have three related problems that I would like to share. They are strikingly similar in nature, but they properly belong in three different branches of mathematics: the first problem is about finite combinatorics, the second [...]

Math hangout: brainswap mayhem

Samuel Coskey — Fri, 17 Feb 2012 03:26:51 +0000

Recently on a social network, Dana Ernst pointed out something called the “futurama theorem”: the characters all swap their brains and need to prove a theorem before they can swap back. It sounded cute, so I made it the topic of a hangout discussion.

Amy swaps brains with mop bucket

Of course, it would normally be very easy to swap back, just retrace your steps, but the show introduced a (silly of course) technical hurdle: once a pair of bodies have had their brains swapped, they can never have their brains swapped again. (The reason they give has something to do with the body’s immune system rejecting a second swap.)

This hurdle is quite serious: the only way for two people who have swapped their brains to ever get back to normal is to bring some new people into the story! So initially, we must ask:

Suppose that a group $A_1,\ldots,A_n$ of people have had their brains mixed up in some fashion. Can we bring in some number new people $B_1,\ldots,B_m$ (who are “clean” in the sense that they never swapped) and use them to restore $A_1,\ldots,A_n$ to their original brains?

The answer is “yes”; in fact it is not hard to see that $n$ clean people will suffice. You simply dump all of the brains in bodies $A_1,\ldots,A_n$ into the bodies of $B_1,\ldots,B_n$, and then put them back one by one into the correct place. (Of course, if some $A_7$ already had his correct brain, then you actually don’t move him at all, or else you could never put him back this way!) This unfortunately has the adverse effect of leaving $B_1,\ldots,B_n$ all messed up, but this is no problem, because they have never swapped with each other before and so they can simply swap amongst themselves.

The sharper question now becomes:

How few clean people can we get away with?

We quickly saw that just one clean person $B$ is not enough, even if there is only one swapped pair $A_1$ and $A_2$. Indeed, in this case there are only two moves available: swap $B$ with the first body, and swap $B$ with the second body. But if you try applying these moves in either order you will see that they don’t get you back to normal!

Now, in the case of just one swapped pair, we already know that two clean people $B_1$ and $B_2$ is enough. So the next interesting case is to decide whether two clean people is enough to restore three permuted people. To do this, we needed to introduce some notation. Suppose that $A_1,A_2,A_3$, have permuted so that $A_2$ is in $A_1$’s body, $A_3$ is in $A_2$’s body, and $A_1$ is in $A_3$’s body. We will denote this by \[A_2A_3A_1\] In other words, the bodies are the positions and the symbols $A_1$ and $A_2$ are the brains, so to say that $A_2$ is in the first position is to say that the brain of $A_2$ is in the first body, and so on.

Now, if clean people $B_1$ and $B_2$ come along, let us say their bodies are in the fourth and fifth positions, so we write \[A_2A_3A_1B_1B_2\] Then the solution looks like this:
\[B_1A_3A_1A_2B_2\] \[B_1B_2A_1A_2A_3\] \[B_1A_2A_1B_2A_3\] \[B_1A_2A_3B_2A_1\] \[A_1A_2A_3B_2B_1\]
Yay, we did it! Of course $B_1$ and $B_2$ are now swapped, but that’s no problem because they can always just swap back.

Next, we found out that a similar strategy using just two clean people $B_1$ and $B_2$ works for an arbitrary “cycle”, that is, any conundrum of the form $A_2A_3\cdots A_nA_1$. The argument we gave was very ad-hoc in style, so I will not attempt to describe it in this post. We simply saw a pattern and convinced ourselves rigorously that it must work for any such cycle. Then we ran out of time!

Afterward

First, the argument that we gave can be expressed in a much more compact way if you use traditional permutation notation from group theory. This is what you will find if you search for “futurama theorem” and read the first results. Perhaps a more detailed discussion would have led us to “discover” this notation out of a need to express our arguments.

And second, we did not have time to address arbitrary (non-cyclic) permutations. In fact, we were almost done since any permutation can be decomposed into a product of disjoint cycles. If we had more time, this would have been an awesome opportunity to “discover” this beautiful fact.

A large cardinal in the constructible universe

Assaf Rinot — Thu, 16 Feb 2012 16:14:12 +0000

In this post, we shall provide a proof of Silver’s theorem that the Erdos caridnal $\kappa(\omega)$ relativizes to Godel’s constructible universe.

First, recall some definitions. Given a function $f:[\kappa]^{<\omega}\rightarrow \mu$, we say that $I\subseteq\kappa$ is a set of indiscernibles for $f$ iff for every $n<\omega$ and every $x,y\in [I]^n$, we have $f(x)=f(y)$. Let $\kappa\rightarrow(\alpha)^{<\omega}_\mu$ denote the assertion that every function $f:[\kappa]^{<\omega}\rightarrow \mu$ admits a set of indiscernibles of order-type $\alpha$. Then, one defines the $\alpha_{th}$-Erdos cardinal, as follows:
$\kappa(\alpha)$ is the least cardinal $\kappa$ (if exists) for which $\kappa\rightarrow(\alpha)^{<\omega}_2$.
We mention that Erdos and Hajnal proved that $\alpha<\beta$ entails $\kappa(\alpha)<\kappa(\beta)$, and that a cardinal $\kappa$ is said to be Ramsey iff it is the $\kappa_{th}$-Erdos cardinal, i.e., $\kappa(\kappa)=\kappa$.

Silver’s idea for relativizing $\kappa(\omega)$ goes through a reformulation of the statement concerning the existence of a set of indiscenribles, in terms of the failure of well-foundedness of a related poset. This leads us to revisiting this subject:

Definition. A poset $(P,\le)$ is said to be well-founded iff every non-empty subset $A\subseteq P$ has a $\le$-minimal element $x$. (the latter means that $y\le x\rightarrow y=x$ for all $y\in A$)

Fact (ZF). Suppose that $(P,\le)$ is a poset whose underlying set $P$ admits a well-ordering $\unlhd$, then all of the following are equivalent:

$(P,\le)$ is well-founded;
if $f:\omega\rightarrow P$ is an order-reversing map, then its image is finite (slang: “$P$ does not admit an infinite $\le$-descending chain”);
there exists an ordinal $\theta$ and a function $\rho:P\rightarrow\theta$ such that for every distinct elements $x,y\in P$: $x\le y$ implies $\rho(x)< \rho(y)$.

Proof. $(1\Rightarrow 3)$ Recursively define

$R_0:=\emptyset$;
$R_{\alpha+1}:=\{ x\in P\mid \forall y\in P\setminus\{x\}(y\le x\rightarrow y\in R_\alpha)\}$;
$R_\alpha:=\bigcup_{\beta<\alpha}R_\beta$ for all limit $\alpha$.

Of course, $\{ R_\alpha\mid \alpha\in On\}$ is an increasing chain of subsets of $P$, and so by the axiom of replacement, there exists some $\theta\in On$ such that $R_{\theta+1}=R_\theta$. Note that $R_\theta=P$, because otherwise, $P\setminus R_\theta$ would be non-empty, and one could find a minimal $x\in P\setminus R_\theta$; it will then follow from the minimality of $x$ in $P\setminus R_\theta$, that $y<x\rightarrow y\in P_\theta$. So $x\in R_{\theta+1}\setminus R_\theta$, contradicting the definition of $\theta$.
Thus, define (the rank function) $\rho:P\rightarrow\theta$ by letting $\rho(x):=\min\{ \alpha<\theta\mid x\in R_{\alpha+1}\}$ for all $x\in P$. Then $\rho$ has the desired properties.

$(3\Rightarrow 2)$ Let $\rho:P\rightarrow\theta$ witness (3). Suppose that $f:\omega\rightarrow P$ is order-reversing. Then $(\rho\circ f):\omega\rightarrow\theta$ is an order-reversing map from ordinal to ordinals, and hence $(\rho\circ f)[\omega]$ is finite. Since $f[\omega]$ is linearly-ordered by $\le$, $(\rho\circ f)[\omega]$ is finite iff $f[\omega]$ is finite. Consequently, the image of $f$ is indeed finite.

$(\neg1\Rightarrow \neg2)$ Suppose that $A$ is non-empty subset of $P$ without a minimal element. Define a relation $R$ of $\le$ over $A$, as follows: $yRx$ iff $x=\min_{\unlhd}(y_{\downarrow})$, that is, $$x=\min_{\unlhd}\{z\in A\setminus\{y\} \mid z\le y\}.$$

Note that $yRx$ entails $x\le y$, and that for every $y\in A$, there exists a unique $x\in A$ such that $yRx$. Let us say that a function $f$ is good, if $\text{dom}(f)$ is a positive integer, $f(0)=\min_{\unlhd}A$, and if $n+1\in\text{dom}(f)$, then $f(n)Rf(n+1)$.
Let $\mathcal F=\{ f\in{}^{<\omega}P\mid f\text{ is good}\}$. Then a second look at the definition of $R$ reveals that $f:=\bigcup\mathcal F$ is a function. Moreover, since $A$ does not contain a minimal element, we get that $\text{Im}(f)$ is infinite. As $yRx$ implies $x\le y$, we conclude that $f:\omega\rightarrow P$ is an order-reversing function with an infinite image. $\square$

Proposition (ZF). ${}^{<\omega}\kappa$ admits a well-ordering.
Proof. Given $x,y\in{}^{<\omega}\kappa$, we let $x\unlhd y$ iff one of the following holds:

$x=y$;
$|x|<|y|$;
$|x|=|y|$ and $x\neq y$ and $x(\Delta(x,y))<y(\Delta(x,y))$. (here, $\Delta(x,y):=\min\{ n<\omega\mid x(n)\not=y(n)\}$.)

We need to show that $({}^{<\omega}\kappa,\unlhd)$ is linearly-ordered, and well-founded. It is obvious that for every $x,y\in{}^{<\omega}\kappa$, we either have $x\unlhd y$ or $y\unlhd x$, and that if $x\unlhd y$ and $y\unlhd x$, then $x=y$. Next, suppose that $x\unlhd y$ and $y\unlhd z$. We need to verify that $x\unlhd z$. To avoid trivialities, we may assume that $|\{ x,y,z\}|=3$. If $|x|<|y|$ or $|y|<|z|$, then $|x|<|z|$, and we are done. So, we assume that $|x|=|y|=|z|$. Put $n_0:=\Delta(x,y), n_1:=\Delta(y,z)$.

If $n_0\le n_1$, then $x(n_0)<y(n_0)\le z(n_0)$ and $x\restriction n_0=y\restriction n_0=z\restriction n_0$. So $\Delta(x,z)=n_0$ and $x(\Delta(x,z))<z(\Delta(x,z))$.
If $n_1<n_0$, then $x(n_1)=y(n_1)<z(n_1)$ and $x\restriction n_1=y\restriction n_1=z\restriction n_1$. So $\Delta(x,z)=n_1$ and $x(\Delta(x,z))<z(\Delta(x,z))$.

Next, we verify well-foundedness. Suppose that $A$ is a non-empty subset of ${}^{<\omega}\kappa$. If $\emptyset\in A$, then it is the minimal element of $A$, and we are done. Suppose now that this is not the case, and let $n$ be the least natural number for which there exists $x\in A$ with $|x|=n+1$. Put $B:=\{ x\in A\mid |x|=n+1\}$. Next, let $k_0:=\min\{ x(0)\mid x\in B\}$, and put $B_0:=\{ x\in B\mid x(0)=k_0\}$. Then for $i<n$, let $k_{i+1}:=\min\{ x(i+1)\mid x\in B_i\}$, and put $B_{i+1}:=\{ x\in B_i \mid x(i+1)=k_{i+1}\}$. Then $B_{n}$ is a singleton whose element is the minimal element of $A$. $\square$

Terminology. Let us say that two functions $g,h$ are compatible iff for every $i,j\in\text{dom}(g)\cap\text{dom}(h)$, we have $$g(i)\in g(j)\text{ iff }h(i)\in h(j).$$

Lemma (ZF). Given a function $f:[\kappa]^{<\omega}\rightarrow X$, a (countable) ordinal $\alpha$, and bijection $g:\omega\leftrightarrow\alpha$, denote $$S(f,g):=\{ h\in{}^{<\omega}\kappa\mid g,h\text{ are compatible, and }Im(h)\text{ are indiscenrnibles for }f\}.$$ Then $f$ has a set of indiscernibles of order-type $\alpha$ iff $(S(f,g),\supseteq)$ is not well-founded.

Proof. As ${}^{<\omega}\kappa$ admits a well-ordering, we know that $(S(f,g),\supseteq)$ is well-founded iff it does not admit an infinite $\supseteq$-descending chain iff if does not admit an infinite $\subseteq$-increasing chain.
$\Rightarrow$ Suppose that $f$ has a set of indiscernibles of order-type $\alpha$. Fix an order-preserving injection $i:\alpha\rightarrow\kappa$ such that $i“\alpha$ is a set of indiscernibles for $f$. Define $h:\omega\rightarrow\kappa$ by letting $h(n)=i(g(n))$ for all $n<\omega$. As $i$ is order-preserving, we get that $g$ and $h$ are compatible, and then $h\restriction n\in S(f,g)$ for all $n<\omega$. As $\{ h\restriction n\mid n<\omega\}$ is an increasing $\subseteq$-chain, we infer that $(S(f,g),\supseteq)$ is not well-founded.
$\Leftarrow$ Suppose that $(S(f,g),\supseteq)$ is not well-founded, and let $\langle h_n\mid n<\omega\rangle$ be an increasing $\subseteq$-chain within $S(f,g)$. Let $h:=\bigcup_{n<\omega}h_n$. Then $\text{dom}(h)=\omega$, and $g,h$ are compatible. Put $I:=Im(h)$. Then $I$ is a set of indiscerinbles for $f$. Finally, since $g,h$ are compatible and $\text{dom}(h)=\text{dom}(g)$, we get that $Im(g)$ and $Im(h)$ has the same order-type. That is $\text{otp}(I)=\alpha$. $\square$

We know that $\Delta_0$ properties (e.g., being an ordinal, being a function) are absolute for all transitive models of $ZF$, and hence $\Sigma_1$ properties are upward absolute, and $\Pi_1$ properties are downward absolute. In particular, $\Delta_1$ properties are absolute. Let us point out that this is indeed the case here:

Suppose that $f,\alpha,g$ are as in the preceding lemma. Then $f$ has a set of indiscernibles of order-type $\alpha$ iff the following $\Pi_1$ formula (with a parameter $S:=S(f,g)$) fails:

$$\forall A[((A\neq\emptyset)\wedge(A\subseteq S))\rightarrow(\exists x\in A\forall y\in A(y\supseteq x\rightarrow y=x))].$$

Recalling the fact that we proved at the beginning of this post, we get as well that $f$ has a set of indiscerinbles of order-type $\alpha$ iff the following $\Sigma_1$ formula (with a parameter $S:=S(f,g)$) fails:

$$\exists\theta\exists\rho[(\theta\text{ is an ordinal})\wedge(\rho:S\rightarrow\theta\text{ is a function})\wedge(\forall x\in S\forall y\in S(x\supsetneq y\rightarrow \rho(x)\in\rho(y))].$$

[Note that the negation of a $\Delta_1$ statement is again $\Delta_1$]

Corollary (Silver, 1970). Suppose that $\mathcal M$ is transitive model of ZF, and $\alpha,\kappa,\mu\in\mathcal M$ with $\alpha<(\omega_1)^{\mathcal M}$. If $\kappa\rightarrow(\alpha)_\mu^{<\omega}$ holds (in the universe), then $$\mathcal M\models \kappa\rightarrow(\alpha)_\mu^{<\omega}.$$
In particular, if $\kappa(\omega)$ exists, say it is $\kappa$, then $L\models \kappa(\omega)\text{ exists, and }\kappa(\omega)=\kappa$.
Proof. Suppose that $f:[\kappa]^{<\omega}\rightarrow\mu$ is a given coloring in $\mathcal M$, and $\alpha<(\omega_1)^{\mathcal M}$. Our goal is to find (within $\mathcal M$) a set of indiscernibles for $f$ of order-type $\alpha$.
Fix in $\mathcal M$, a bijection $g:\omega\leftrightarrow\alpha$. As $\kappa\rightarrow(\alpha)_\mu^{<\omega}$ holds in the universe, there exists a set of indiscernibles for $f$ of order-type $\alpha$. So $S(f,g)$ is not well-founded. Since $f,g\in\mathcal M$, we infer that $S(f,g)\in\mathcal M$, and so by absoluteness for $\Delta_1$ properties: $$\mathcal M\models S(f,g)\text{ is not well-founded}.$$ It now follows from the lemma that $\mathcal M$ contains a set of indiscernibles for $f$ of order-type $\alpha$. $\square$

For completeness, we mention that Rowbottom proved in his dissertation that the existence of the Erdos cardinal $\kappa(\omega_1)$ contradicts the axiom $V=L$. (Thus, improving Scott’s famous theorem that the existence of a measurable cardinal implies $V\neq L$.) It follows that Silver’s theorem that $\kappa(\alpha)$ relativizes to $L$ for all $\alpha<(\omega_1)^L$ is best possible.

Computable roots of computable functions

Carl Mummert — Tue, 14 Feb 2012 02:06:03 +0000

Here are several interesting results from computable analysis: Theorem 1. If $f$ is a computable function from $\mathbb{R}$ to $\mathbb{R}$ and $\alpha$ is an isolated root of $f$, then $\alpha$ is computable. Corollary 2. If $p(x)$ is a polynomial over … Continue reading →

The Delta-System Lemma

Micheal Pawliuk — Sun, 12 Feb 2012 22:38:05 +0000

This is a delta system of rivers. I didn't know this until last year.

Now that you know the basics of countable elementary submodels (CESM), you might think that you are in the clear. “Mike”, you say arrogantly, “I know all the most basic properties of CESMs, without proof I remind you, what else could I possibly want?”. I gently and patiently remind you that CESMs are worthless unless you know how to apply them properly.

So let’s do that.

Here are two theorems whose proofs you might already know, but that can be proved using elementary submodels. I will show you a proof of the $\Delta$-system lemma (a fundamental lemma in infinitary combinatorics) and a topological theorem of Arhangel’skii. Both of these proofs are taken from Just & Weese’s book “Discovering Modern Set Theory 2”, chapter 24.

The $\Delta$-system lemma.

This lemma is an extremely useful tool for dealing with uncountable families. It is the tool most often used to show that a partial order has the countable chain condition.

(The $\Delta$-system lemma) Let $B \sse [\omega_1]^{<\omega}$ with $B$ uncountable. Then there is an uncountable $A \sse B$, and $r \in [\omega_1]^{<\omega}$ such that $\forall a,b \in A, a\cap b = r$.

Here $r$ is called the root, which is a (possibly empty) finite subset of $\omega_1$. The lemma says we can refine any uncountable family of finite subsets of $\omega_1$ so that the elements of the family agree precisely on the root.

Note that the “countable $\Delta$-system lemma” is false. Letting $B := \{\{0,1,2,3, \dots, n\}: n\in \N \}$ is a countable subset of $\omega_1$ such that no finite subset of $\omega_1$ could possibly be a root for any refinement of $B$.

Proof of the $\Delta$-system lemma using CESM. Ok, so we are actually going to show something a little bit stronger.

(Lemma) Let $B \sse [\omega_1]^{<\omega}$ with $B$ uncountable. There is a countable $N \sse \omega_1$ such that every $b \in B$, with $b \not \sse N$ is contained in an uncountable $\Delta$-system $A \sse B$ with root $r \sse N$.

(Whoa, what is this saying?)

(Well it is using the fact that we don’t need to concern ourselves with the countably many finite subsets of $N$, as we can throw those away in the refining process. Now the lemma says that $B$ can be refined to a $\Delta$-system. What about this condition that $N$ is a countable subset of $\omega_1$? It doesn’t really matter.)

To the proof of the lemma!

Let $M$ be a CESM of set theory with $B\in M$. Now let $N := M \cap \omega_1$, which we know is a countable ordinal, $\delta$. This $N = \delta$ will turn out to satisfy the conditions of the lemma.

Note that $\exists b\in B$ such that $b \not\subseteq N$, because $B$ is uncountable but $N$ is countable. We define $r:= b \cap N$, and this will turn out to be the root of an uncountable $\Delta$-system that is contained in $B$.
So for any given $\alpha<\delta$, $$V \models r \sse b \wedge b \setminus r \neq \emptyset \wedge min(b\setminus r) > \alpha$$

In order to use the elementarity of $M$ we change that statement to include an existential quantifier instead of refering to $b$, which isn’t in $M$. Remember that $M$ cannot refer to things that are outside of it: $$V \models \exists x \in B \color{grey}{(r \sse x \wedge x \setminus r \neq \emptyset \wedge min(x\setminus r) > \alpha)}$$
(We use elementarity now)

The statement above references only things in $M$, so by elementarity, we must have: $$M \models \exists x \in B \color{grey}{(r \sse x \wedge x \setminus r \neq \emptyset \wedge min(x\setminus r) > \alpha)}$$

(Magic time)

Remember we let $\alpha < \delta$ be arbitrary so: $$M \models \forall \alpha < \delta (\exists x \in B) \color{grey}{(r \sse x \wedge x \setminus r \neq \emptyset \wedge min(x\setminus r) > \alpha)}$$

(Yeah, so what?)

Well recall that $\delta = M \cap \omega_1$, so infact: $$M \models \forall \alpha < \omega_1 (\exists x \in B) \color{grey}{(r \sse x \wedge x \setminus r \neq \emptyset \wedge min(x\setminus r) > \alpha)}$$

So now because $M$ is an elementary submodel of set theory, $$V \models \forall \alpha < \omega_1 (\exists x \in B) \color{grey}{(r \sse x \wedge x \setminus r \neq \emptyset \wedge min(x\setminus r) > \alpha)}$$

So it is true that for any $\alpha<\omega_1$ there is an $x\in B$ such that $r \sse x$, $x \setminus r \neq \emptyset$ and $min(x\setminus r) > \alpha)$. From here we can easily construct an uncountable sequence of elements in $B$ with root $r$. [QED]

Did you notice how important it was that $M \cap \omega_1$ was a countable ordinal? Because of that we were able to find an element of $B$ sufficiently far out in $\omega_1$ that agrees with the root. We did this using only the fact that countable < uncountable.

Nothing fancy at all!

The unsettling part here is that we found this $b$ that was larger than both $\alpha$ and $M$’s copy of $\omega_1$ which was $\delta$. Then elementarity said: “Finding a $b$ further than $\alpha$ was the important part, I don’t care that it was bigger than $M$’s $\omega_1$ as long as this $b$ is less than the real $\omega_1$; I’ll take care of making it less than $M$’s $\omega_1$”. We did that, but we ended up only finding b’s that were greater than countably many $\alpha$. Elementarity jumps in with “Don’t worry, checking it for the countably many $\alpha< \delta$ is fine, I’ll make sure it works for all $\alpha<\omega_1$”.

Thanks Elementarity!

A more standard proof of the $\Delta$-system lemma.

Here is the more ~~elementary~~ basic proof of the lemma: Let $B \sse [\omega_1]^{<\omega}$ with $B$ uncountable. Since $B = \bigcup_{n<\omega} [\omega_1]^n$, there is an $N$ such that $B_1 := B \cap [\omega_1]^N$.

(We reduced to the case where all elements of $B$ have the same length).

We now prove the lemma by induction on $N$.

If $N=1$ it is clear that $B_1$ itself is a $\Delta$-system with root $r = \emptyset$.

Suppose the statement is true for $N=k$. Write $B_1 = \{ A_{\alpha}\cup\{x_\alpha\}: \alpha < \omega_1\}$, with each $A_\alpha$ having cardinality $k$. We would like to apply the induction hypothesis to $B_2 := \{A_\alpha : \alpha<\omega_1\}$, which could only happen if $B_2$ was uncountable. (If it was countable, then there are uncountably many elements of $B_1$ that all share the same $A_\alpha$, then all of the $x_\alpha$ must be different, so we have our uncountable $\Delta$-system which is a subset of $B_1$).

Now suppose $B_2$ is uncountable. Apply the induction hypothesis to $B_2$ to get $B_3 = \{C_\alpha \cup \{y_\alpha\}: \alpha < \omega_1\}$. (Here we are just changing the ‘A’s to ‘C’s and ‘x’s to ‘y’s as we have taken a refinement). Now $\{y_\alpha: \alpha<\omega_1\}$ is either countable (in which there are uncountably many $C_\alpha$ that share the same $y$, so we can add that $y$ to the root).

Otherwise if $\{y_\alpha: \alpha<\omega_1\}$ is uncountable…

You get the idea. The point being that this proof is just a lot of refining and using the pigeonhole principle.

It is interesting to note that Stevo Todorcevic insists that these two proofs of the $\Delta$-system are the “wrong” ones. He is convinced that the right one is the Ramsey-Theory tree proof.

Looks like I’ve written a lot already and will get to the topological proof later.

How not to teach logic

Samuel Coskey — Fri, 10 Feb 2012 05:24:38 +0000

Some faculty at York think it’s ok to omit parentheses. There’s not much to say other than “wow”.

Take a look at the first and most fundamental formulas of predicate calculus, as summarized on a formula sheet for an intorductory logic course:

The first problem is that the author is employing the over-used triple-bar symbol to denote material bi-implication, as opposed to the much more reasonable and popular ↔ symbol. The second problem is that these formulas are so ambiguous that they can’t even be said to be false.

But, I hear you say, one should perhaps assume that the parentheses are implicit, and moreover that they are left-associating by default. So in fact, in the context of the class, these formulas may well be correct! Such an argument, aside from the fact that it presupposes the instructor is brewing up a pedagogical nightmare of a syllabus, is initally somewhat convincing. Until you read a little further down on the page:

Seriously, just look at it.

To be clear, this isn’t the worst of it. The professor who compiled these notes at least included some parentheses, you know, for when things get super-duper-hyper ambiguous. Unfortunately, a number of other professors teach the same course from the same notes. Only when they copy the notes, they don’t tend to include any parentheses (I’ll scan it if I see it again).

What does this say to students? It says, loudly and clearly:

I’m too busy to worry about whether my course materials are universally confusing or not. In fact, figuring this shit out is really your job. If you can’t figure out what I meant, then I don’t see how you could pass this course anyway.

And to that I would add: “Besides, I don’t give two dips about the material, since it doesn’t have anything to do with the way mathematicians think about mathematics anyway. Why are we studying it? Something about maturity. Ask the department chair.”

Update. Martino points out the following: If you insist on a particular set of rules for operator precedence, and if you then insert enough parentheses to make these ambiguous formulas well-formed, then they will be true regardless of which choices you made along the way!

This doesn’t, of course, make the formulas any less confusing. In some ways it makes it more confusing: that several axioms (some of which are more perverse than others) are being combined into one without comment.

If you build it, will they come?

Peter Krautzberger — Tue, 07 Feb 2012 04:08:49 +0000

Last week, Tim Gowers published another post on the publishing debate entitled “Abstract thoughts about online review systems”.

There are, as usual, interesting thoughts both in the post and in the comments (including some very nice elitist ones), enough …

An AIM workshop on set theory and C*-algebras

Samuel Coskey — Wed, 01 Feb 2012 05:16:29 +0000

Last week I attended a workshop on set theory and C*-algebras at the AIM. Here is how it went…

When you approach the American Institute of Mathematics, you get the distinct impression that you are about to see some seriously low, low prices on mp3 players. The reason for this is that you cannot approach the AIM without approaching a Fry’s Electronics; a discount superstore with a poor ethical history. In fact, the AIM is behind an ill-marked door leading to a windowless cave adjacent to the store. It is a dry office space, filled with books and binders full of dry leaves of aging mathematics. And there is coffee and bagels for breakfast.

Founded by a Mr. John Fry some 20 years ago, the AIM hosts a special workshop in mathematics each week. The format diverges, however, from the usual endless barrage of talks. Instead, there are just a few introductory talks followed by clarification sessions. Later in the week, the questions transition into research questions, with voting to determine which ones to attack, also in small sessions. There are also impromptu talks of a specialized nature when there is sufficient demand.

Last week, the topic was the overlap of set theory and C*-algebras, which is one of my (lately) favorite areas. During the introductory portion of the workshop, I was glad to have the opportunity to become more comfortable working with C*-algebras in general. However, my questions in particular were of a somewhat basic nature, and I found it hard to get such questions answered. My session was half filled with ringers who would push the speaker onward. In this case, I actually would have preferred a more traditional format; perhaps a well-prepared talk just for the non-C*-algebraists, one which methodically addressed a range of basic facts.

One of the most useless moments for me came when I asked a question about dimension in C*-algebras:

Sam: (some question about nuclear C*-algebras)
Self-appointed moderator: (some techno-babble ending with…) Like take a CW complex.
(pause)
Sam: Are you talking to me?

This was followed by general laughter and then wine. It was not followed by an y hint of what is a CW complex, nor of what the heck it has to do with anything said so far this week.

Later in the week, I joined a session which focused on dynamical properties of the automorphism group Aut(A) for some very well behaved algebras A. The goal was to establish one fixed property (namely turbulence, which I’ll hopefully address in a later post), and also to obtain a template for establishing this property for more and more algebras A. Unfortunately, it’s rather difficult to use 13 different humans as problem-solving resources, since for instance they cannot all be kept up to speed all the time. In fact, I was often left behind as the “real” problem-solvers plowed on.

Despite any criticisms, the conference was an overall success for me. First, I did somehow greatly expand my basic understanding of C*-algebras. And second, the problem-solvers failed in the end (ha!), while on the other hand I have very good notes and I can certainly continue working on this and related problems in the future.

I’ll conclude by giving the workshop a letter grade: B

A comment on Tim Gowers’s blog

Peter Krautzberger — Sun, 29 Jan 2012 20:28:15 +0000

I just left an awfully long comment on Tim Gowers’s blog. Thanks, François for mentioning the new post on twitter! ~~Incidentally, it seems that my email is considered spam by Akismet these days, so it will most likely never show~~
…

Groups in $\beta \mathbb{N}$

Peter Krautzberger — Thu, 26 Jan 2012 05:07:16 +0000

Well, if you haven’t read my recent introspection, consider yourself warned: the video below is, in my humble opinion, a rather poor talk I gave at our Logic Seminar here at the University of Michigan.

Nevertheless, I would like …

A Practical Guide to Using Countable Elementary Submodels

Micheal Pawliuk — Thu, 26 Jan 2012 02:24:41 +0000

Elementary submodels are like dollhouses in your real house

So, you may have heard about these things called countable elementary submodels. You may have heard that they work like magic and do all sorts of amazing things. “Mathematical voodoo” some might say. “Witchcraft!” others declare. Hearing this you become intrigued and set out to harness this black power. You quickly realize that there are very few places to learn this dark art; the protectors of this knowledge don’t want it leaking out.

Here I hope to lay out the essential things you need to know (and omit the things you don’t need to know) so that you can start using countable elementary submodels. I am going to lay out as little of the machinery as possible and display only the relevant applicable facts you will need for most proofs involving elementary submodels.

1. A Countable Submodel of What?

The universe of all sets $V$ is a ‘model’ for set theory, but it is too big. If we did have a model for set theory we would know that there is a countable submodel of it, by Lowenheim-Skolem. Of course we can’t assert that set theory has a model as this would be equivalent to asserting the consistency of set theory. The clever way around this is to realize that any proof in mathematics only ever uses finitely many axioms of set theory and references only finitely many specific sets. It is always possible to find a model $H$ of those finitely many axioms and special sets. (Aside, for those of you who have seen this before, why doesn’t this violate the compactness theorem? It’s tricky.) Here $H$ will be our copy of the universe, just for a given proof, and we will take a countable submodel of $H$, not $V$. This is where the language “Take a large enough fragment of ZFC” comes from.

As it turns out there is a class of sets that we usually draw $H$ from. We usually take $H$ to be a set $H(\alpha)$, where $\alpha$ is a cardinal and $H(\alpha)$ is the set of all sets hereditarily of cardinality less than $\alpha$. This doesn’t really matter at all. So don’t fret about this.

2. What does the word ‘elementary’ in CESM mean?

A submodel $M$ of $H$ is an elementary submodel (denoted $M \prec H$) if any statement true in $H$ is true in $M$ (given that we have relatavized the statement to $M$). Basically, if $H$ and $M$ have the same language and we can express a statement in that language it should be true in $H$ iff it is true in $M$. Relativizing a statement to $M$ means that we only quantify over things in $M$, not all of $H$. Also, we are not allowed to reference things outside of $M$.

For example, as dense linear orders $M := \Q \cap (0, \infty) \prec \Q$. We know that in $\Q$ the following statement $\phi$ is true: $$(\forall x)( \exists y )[x<y]$$ Well what convention did we just use? We just assumed that the $\forall$ only quantifies over rational numbers (and not say over complex numbers, or sets). Relatavizing this statement to $\Q \cap (0, \infty)$ we see that it is also true, and now the statement $\phi^M$ is $$(\forall x \in \Q \cap (0, \infty))( \exists y \in \Q \cap (0, \infty))[x<y]$$ Notationally we say $\Q \models \phi$ and $\Q \cap (0, \infty) \models \phi$. (See Chris Eagle’s comment below)

Basically relativizing makes sure that a model can actually say something about the statement.

3. What does it mean for a model to “think something is true”?

Saying that $M$ thinks that a statement $\phi$ is true is simply shorthand for $M \models \phi$. So what? First of all we get paradoxical conclusions like $M \models 2^\omega$ is uncountable. How can this be because $M$ is countable?

Well, a model thinks that $ 2^\omega$ is uncountable because all of the injective functions $f: \N \rightarrow 2^\omega$ that are in $M$ are not onto. So even though $M$ actually has only countably many elements of $ 2 ^\omega$, the model also doesn’t have any way to check that. Even though from the outside we see that there is a bijection $f: \N \rightarrow 2^\omega$, this function is not in $M$.

The thing to be careful about here is that $M$’s copy of $2^\omega$ is not the real copy of $2^\omega$. (How could it be?! $M$’s copy has only countably many elements!)

4. How do I determine when something is in $M$?

SUPER USEFUL FACT: Any set definable from parameters in $M$ is actually a member of $M$.

Remember that sets in $M$ sometimes have the form $\{x \in M : \phi(x_1, x_2)\}$ where $x_1, x_2$ are free variables in the statement $\phi$. If the objects in $\phi$ are all in $M$, then $\phi$ is definable from parameters in $M$.

For example, back in $\Q \cap (0, \infty) \prec \Q$, $\{x : x<-1\}$ is definable in $\Q$, but not in $\Q \cap (0, \infty)$.

This is kind of like super useful fact from forcing that asserts that I am allowed to define sets in the ground model even if I use the forcing symbol in definition, so long as all the other parameters are in the ground model.

5. What is the big difference between $A \in M$ and $A \sse M$?

During a proof using CESM you will be using the super useful fact over and over. Why does it matter? Think about how sets work with respect to set operations (Power set, intersection, union, image of a function, etc.). In that case you want to know that applying these operations to a set will produce a set. In symbols, $A\in V$ asserts that $A$ is a set, but $A \sse V$ does not assert that. For example the class of all ordinals ON $“\sse” V$ and is not a set, so we can’t do things like take $\P(\textbf{ON})$.

If $A \in M$ then the model “knows about A” and we can use the super useful fact.

6. What is the really important thing I need to know about finite and countable subsets of $M$?

For sets of small cardinality sometimes we automatically get information about them.

FACT: Any finite subset of $M$ is actually a member of $M$.

FACT: Any countable member of $M$ is actually a subset of $M$.

7. What are the really important things I need to know about ordinals?

Many theorems using CESM start of by considering $M \cap \omega_1 := \delta$.

FACT 1: $\delta$ is a countable ordinal. That is, $\delta \in \omega_1$.

FACT 2: If $A\in M$, $A\sse \omega_1$ and $\delta \in A$ then $A$ is a stationary set in $\omega_1$. In particular, $A$ is uncountable.

Moreover we get the folowing (quite general) fact:

Assaf’s FACT: If $A\in M$ and there exists some $x\in A$ which is not in $M$, then $A$ is uncountable.

Proof. Recall that any countable $A\in M$ is also a subset of $M$.

What else?

This should give you the basics that you need to know in order to read a proof using CESM. For more information about CESM see chapter 24 in Just & Weese’s book “Discovering Modern Set Theory; Volume 2″. I hope that this was helpful and please let me know if there is anything that needs changing.

Next week I will tackle two proofs using CESM.

Mostly, I’m a terrible speaker

Peter Krautzberger — Wed, 25 Jan 2012 01:40:31 +0000

They say, ignorance is bliss.

As I wrote in a comment recently, almost all mathematicians are bad speakers for almost all audiences. It seems to me that most people would agree. But in blissful ignorance, we don’t worry about our …

Testing…

Dana Ernst — Mon, 23 Jan 2012 22:20:04 +0000

This is just a placeholder.

Live blogging from #scio12

Peter Krautzberger — Thu, 19 Jan 2012 15:42:47 +0000

So, I will attempt to live blog from Science Online 2012. This means this post will get updated instead of writing many posts.

Wednesday evening

Getting to RDU was easy, direct flight, all good. Once there I hooked up …

An iPhone app for a nonstandard model of number theory?

Victoria Gitman — Wed, 18 Jan 2012 18:49:09 +0000

This is a talk at the City Tech C-LAC (Center for Logic, Algebra, and Computation) Seminar, February 14, 2012.

A rigorous study of number theory, the properties of the natural numbers under the operations of addition and multiplication and the usual ordering, began more than two millenia ago in ancient Greece.

Giuseppe Peano

But it was only in the late 19th century that the first explicit axiomatization for it was proposed by Giuseppe Peano. Reinterpreted in first-order logic, the generally accepted language of formal mathematics formulated in the early 20th century by Thoralf Skolem, the Peano axioms consisted of two lists. The first finite list summarized the fundamental properties of addition, multiplication, and the ordering. The second infinite list consisted of induction axioms for every first-order definable number theoretic property. Because first-order logic does not allow quantification over properties, expressing induction required an infinite list. Could anything else besides the natural numbers possibly satisfy the Peano axioms? The answer came from several fundamental results about first-order logic established in the 1930′s. In 1931, Kurt Gödel proved the first incompleteness theorem showing that there is a true number theoretic property that is not provable from the Peano axioms. A model of the Peano axioms together with the negation of such a property had to be different from the natural numbers. An even stranger result already followed from Gödel’s countable compactness theorem (1930) showing that any collection of first-order properties in a countable language, whose every finite subset has a model, has a countable model. Using the countable compactness theorem, we can show that for every infinite binary sequence $s$, there is a countable model of ALL true number theoretic properties that has an element that is divisible by the $n^{\text{th}}$ prime number if and only if the $n^{\text{th}}$ digit of $s$ is $1$. Yes, there are models of number theory with elements divisible by infinitely many primes! Finally, once Anatoly Maltsev proved the uncountable version of the compactness theorem, it followed that there are models of the Peano axioms (even of all true number theoretic properties) of every infinite cardinality. Unintuitive as it may seem, there are uncountable models of number theory. We call the natural numbers the standard model of the Peano axioms and all these other models nonstandard.

Wouldn’t it be nice to actually construct a nonstandard model of number theory? In other words, does there exist an algorithms to compute the addition and multiplication of some such model? Well, first off, does there exist an algorithm to compute the addition and multiplication on the natural numbers? Of course, without it there would be no debate in education circles about the use of calculators in the classroom. And, of course, there is an iPhone app for it as well. Almost. A calculator or an iPhone app or any other computing device can only add and multiply numbers up to whatever size is allowed by its memory capacity. But let’s all be theoreticians here and suppose that our computing devices do have infinite memory as in the Turing model. So can we have an iPhone app for a nonstandard model of the Peano axioms (allowing for infinite memory)? Is the question even meaningful? Computers manipulate numbers and elements of a nonstandard models are not numbers. But that is easily addressed since if the nonstandard model is countable we might as well assume for all intents and purposes that its elements are numbers. After all that’s what allows computers to manipulate all other objects that are not numbers such as text, images, etc. So a more precise formulation of the question becomes: is there a countable nonstandard model of the Peano axioms and a way to associate its elements with the natural numbers so that we can write an algorithm to calculate the nonstandard operations of addition and multiplication for it?

Stanley Tennenbaum

In 1959, Stanley Tennenbaum showed that the answer is NO! It is simply not possible to get our hands on a nonstandard model of the Peano axioms. Put broadly (vaguely) the reason is that any nonstandard model of the Peano axioms codes information that cannot be accessed algorithmically. These creatures are too complex for those limited hands of ours. This is not to imply that we cannot study the properties of these truly fascinating structures, a thriving field of research of which I am a proud member. It is just that we can never have an iPhone app for one.

In this talk, I will outline the background and the proof of Tennenbaum’s theorem. Take a look at the slides here.

Secret Santa 3: The Paradox.

Micheal Pawliuk — Tue, 17 Jan 2012 01:06:36 +0000

Last time I discussed the solution to Sam’s problem:

Sam’s Problem. Is it possible for two people to each choose a natural number so that both numbers are exactly 1 apart and neither person knows who has the larger of the two numbers?

I established, by induction, that it is impossible to do this. Great. We write “QED” and move on. There is a very convincing counter-argument that was brought to my attention by Jacob Tsimerman and a student at the Winter Canadian IMO camp. They proposed a method that seems like it should solve Sam’s problem in the positive. What exactly is going on in their method? Where is the mistake?

Jacob’s method. Player 1 chooses a large enough number N (say greater than 100); this is now their number. Player 1 writes down the numbers N+1 and N-1 on different pieces of paper and presents them face-down to Player 2. Player 2 chooses one of them and burns the other one without looking at it. The number Player 2 sees is their number.

This method seems to satisfy the conditions of the problem; The numbers are certainly exactly 1 apart and it doesn’t seem like they know who has the larger number. If you were to use this method in real life it isn’t clear if you have the larger number – we don’t know anything about the second player’s action, how can we possibly know who has the larger number?!

Well it comes down to the words “chooses a large enough number N”. Player one cannot pick the number 0, partly because player 1 will know they have the smaller number.

(1) So can Player 1 choose the number 1? If they do, Player 2 will draw either the number 0 (in which case they know that the game has been broken, as they have the smaller number) or they draw the number 2. Since Player 2 has not announced that the game is broken, Player 1 must then know that Player 2 has drawn the number 2. So in either case Player 1 cannot have the number 1.

(2) Can Player 1 choose the number 2? If Player 2 draws the number 1, they know that player 1 wrote down the number 2 (and not the number 0), so they know the game is broken. If Player 2 draws the number 3, then because Player 2 has not announced that the game is broken Player 1 will know that Player 2 has drawn the number 3. That means that Player 1 knows that the game is broken. So Player 1 cannot write down the number 2.

It continues on in this fashion where one of the players always knows that the game is broken. That is a bit of a cop out. Try to work out the case for Player 1 writing down 3. There will be a back-and-forth where Player 2 knows the game is broken because Player 1 said nothing after Player 2 said nothing after drawing his number. Fun, right?

I was skeptical at first about this because of the amount of calculations this takes. How long do the players need to wait before they know that the game is broken? At first I thought that the number of calculations grew exponentially with the size of number that Player 1 chooses. It turns out that if Player 1 writes down the number 200, then after 200 or so calculations, one of the players will know that the game is broken.

Practically speaking though, humans are incapable of this type of exact, error-free calculations. So practically speaking Jacob’s method does solve Sam’s problem. What about for computer players? Suppose we enforce that after a specified unit of time (say a yoctosecond, $10^{-24}$ seconds) each computer announces whether or not it knows that the game is broken. It seems that if computer 1 chooses a truly enormous number (say greater than a googol bang bang bang) then there just won’t be enough time in the computers’ lifetimes to discover that the game is broken. It is pretty easy to choose an enormous number, so maybe even then Sam’s problem is practically solvable.

What do you think? Is Jacob’s method a legitimate solution to Sam’s problem, or not?

After the common room, how about the seminar room?

Peter Krautzberger — Thu, 12 Jan 2012 02:50:11 +0000

The most impressive community on the (specifically mathematical) intertubes is MathOverflow. Not only because François is one of the moderators and Joel is the #1 power user or because it has introduced many mathematicians to the unexplored possibilities of …

An incompleteness theorem for βn models

Carl Mummert — Thu, 12 Jan 2012 02:27:46 +0000

My first paper was “An incompleteness theorem for $\beta_n$ models” with Stephen Simpson [1]. It’s a short paper, but the idea is very pretty. We know that the incompleteness theorem implies there are strange models of arithmetic, but these models … Continue reading →

Writing math on the web II

Samuel Coskey — Wed, 11 Jan 2012 02:54:09 +0000

This is the first follow-up to my previous post on the subject.

In this post, I will make last time’s question into a slightly more specific proposal. I will continue to take the naive approach of working within the confines of html and css. I simply haven’t had time to investigate the more advanced internet document authoring technologies that seem to be available.

Here is the current status of my vision. One would write something like this:

<html>
 <head>
 <script type="text/javascript" src="sam's-magic-file.js">
 </script>
</head>
<body>

The script, sam's-magic-file.js will load MathJax, style sheets, and parse any “preamble” commands (discussed below). Lastly, the script will perform one key operation: convert double-newlines into paragraphs, much the way WordPress does. Other than that, I do not want the script to interfere. I would prefer it if as much of the heavy lifting as possible was handled with styles.

Next, the user can begin hacking away in the pseudo-LaTeX that I discussed last time. Here is the start of one of my sample documents (the output is shown at the end of the post).

<section>Ergodic components of Grassmann space</section>

In this section we examine the ergodic components for the homogeneous
space $Gr_k\QQ_p^n$ more thoroughly.  We begin by considering the
following simplified statement of Theorem A:

<theorem id="thm_novi">
  Suppose that $n\geq3$ and let $k,l\leq n$.  Let $\Gamma\actson X$ be
  an ergodic component for $SL_n(\ZZ)\actson Gr_k\QQ_p^n$, and let
  $\Lambda\actson Y$ be an ergodic component for $SL_n(\ZZ)\actson
  Gr_l\QQ_p^n$.  If there exists a permutation group isomorphism
  $\Gamma\actson X\longrightarrow\Lambda\actson Y$, then $l=k$ or
  $l=n-k$.
</theorem>
...

As you can see, the word <html> at the top of the document was a misnomer, because <theorem> is not valid HTML. But the simple fact is that no browser cares. Thus, in this world you are free to use standard HTML directives such as <em>, <ul>, styles, and so on.

My stylesheet does everything it can to typeset this code. Theorems are numbered and preceded with the word “Theorem”, proofs are supplemented with a QED symbol, and so on. There really isn’t much to it!

It is also easy to use the same code as a plug-in. Just load the same script at the top of your file, and specify <div class="the-latex-code-is-here">...</div> to apply the styles and scripts to just a part of your document.

We also have to allow the user to extend the command-set like LaTeX does. For formulas, this is easy because MathJax allows you to specify a preamble:

<script type="math/tex">
  \newcommand{\NN}{\mathbb N}
  \newcommand{\ZZ}{\mathbb Z}
  ...
</script>

I would augment this by allowing the user to specify new theorem environments and a variety of style choices (AMS article vs. book) in a similar environment:

<script type="text/x-latex-config">
  ... ?
</script>

But I haven’t yet decided exactly how these directives should be formatted.

Finally, I must remark that CSS stylesheets aren’t quite “there” yet. For instance, it has been proposed that styles should let you specify printing directives including margins, running headers, and page numbers. But many of these features are not yet available. Here, I would grudgingly allow the use of scripts to fill in the gaps until the W3 finally gets its ideas finalized and implemented (this will be a while). This is another area where there is much to be gained by using XML and its more powerful stylesheets, XSL. However, as far as I know the trade-off would be seamless coexistence with other html content, and that seems like a big loss to me.

Have a look at the sample source and printed output of my current experiments. I haven’t handled references yet. I promise to post the code sometime soon!

AMS/ASL Special Session on Alan Turing

Carl Mummert — Tue, 10 Jan 2012 00:03:14 +0000

I arrived home yesterday from the 2012 Joint Mathematics Meetings in Boston, where I was a co-organizer of the AMS/ASL Special Session on the Life and Legacy of Alan Turing. The talks were wonderful, and the session went very smoothly, … Continue reading →

The Secret Santa Problem (Part 2)

Micheal Pawliuk — Wed, 04 Jan 2012 18:43:35 +0000

Last time, just in time for Christmas, we looked at the Secret Santa Problem. Basically the problem is to set up a secret santa type gift exchange without using any external aids like random number generators. A similar problem given to me by Sam Coskey is the following:

Sam’s Problem. Is it possible for two people to each choose a natural number so that both numbers are exactly 1 apart and neither person knows who has the larger of the two numbers?

When Sam gave the problem to me he intended that each player would choose a natural number and then they would sequentially ask questions to each other, possibly refining their original numbers. In this sense it is more like a game of Guess Who than secret santa.

After much thinking, it turns out that there is a fairly easy solution to this problem.

Part 1. Can either player choose the number 0?

Well no, because that person would know that they have a smaller number than the other player.

Part 2. If both players know that $1, 2, \dots, k$ cannot be chosen then $k+1$ cannot be chosen (as $k+1$ would have to be the smaller of the two numbers). So by induction, no number can be chosen by either player.

The lesson here is that induction is a very useful technique! This sounds naive but, when problem solving for contests, induction is often overlooked. Here is another related problem:

Father/Son problem. Is it possible for two players to each choose a human being so that the two humans are father and son, but neither player knows who has the father and who has the son?

The solution is again fairly simple, and uses induction. This time we need a different type of induction. We observe that no player can pick the first ever human being (as they would have the father of the other player’s choice). Now if there is a set S of humans that cannot be chosen, then the sons of people in S cannot be chosen either.

There you go, induction wins again!

Introducing Cantor’s Attic

Victoria Gitman — Tue, 03 Jan 2012 19:55:23 +0000

Two months ago, Joel Hamkins approached me about collaborating on creating a wiki for notions of infinity of all breeds and scales. Joel said that his motivation came from the excellent Complexity Zoo wiki started by Scott Aaronson (the zookeeper) that catalogs notions of computational complexity under the animal zoo metaphor. Our attic metaphor, with the suggested scenery of endlessly ascending staircases, came out of a brainstorming email exchange with Peter Krautzberger and Sam Coskey. The attic had to be Cantor’s, that first serious student of infinity. After some more debate, we settled on MediaWiki, the software behind Wikipedia, as the repository for the site. Now Cantor’s Attic is live and according to Joel “out of the embarrassing stage”. The wiki content is subdivided into four attic levels and a parlour. The upper attic contains large cardinal notions, the infinities whose existence is independent of the ${\rm ZFC}$ axioms. The middle attic contains those infinities whose existence is provable in ${\rm ZFC}$. The lower attic contains remarkable countable ordinals, and the parlour, aspiring to but not quite making it to the infinite, contains very large finite numbers. Joel has populated many entries with information in his characteristicly thorough fashion. The rest will have to be done by the army of volunteers we are hoping to recruit and/or coerce into service. I am apparently responsible for the Ramsey cardinals family, but have instead been bogged down by technical issues. The technical side of setting up the wiki is still a work in progress. The MathJax extension is up and running, the install was a relatively trivial affair. In the process of figuring out how to shorten the page urls by eliminating the ugliness that is GET variables, I learned about a wonderful tool called .htaccess files and mod rewrite rules (here is an excellent tutorial). Next, Joel tried to decorate the site with those endless staircases only to discover that creating thumbnails generated an error of unknown origin. After trolling 3 google pages worth of forums, the issue turned out to be a limit on shell memory, imposed by MediaWiki, that needed to be extended for Imagemagick, the image resizing software, to run (see here for a resolution). Go figure.

Descent into History photo by Ernie Reyes

After that we tackled finding a reference extension that supported bibtex code. A quick search turned up the Biblio extension, which is apparently the most widely used MediaWiki reference extension, but does not support bibtex, and the Bibtex extension, which does, but did not have all the desirable features of Biblio, such as storing your references in a separate “database” page. The obvious thing to do was to combine the two extensions, a task I was not looking forward to, and was saved from by finding NMR Wiki that already did the job with the Biblio-Bibtex extension. The extension is great and my only gripe is that the bibtex code, should you want to see it, appears in a popup window (this comes from the Bibtex extension). Project: change the javascript to have it work as in the Papercite plugin for WordPress. Order of the day: set up the “References” page on the wiki and start adding references. For future projects: see the Community Portal and Discussion pages. Follow-up post: editor access and spam. Meanwhile, we hope our fellow boolesringers will contribute to Cantor’s Attic and help out with technical criticisms/suggestions.

Representing Booles’ Rings at Science Online 2012

Peter Krautzberger — Mon, 26 Dec 2011 23:30:12 +0000

At the end of January, I’ll be at Scio12! Science Online is that super awesomesauce (un)conference where all the hipster science bloggers hang out — and also those fantastic little companies working on the future of doing science on …

Hanging out with Sam

Peter Krautzberger — Thu, 22 Dec 2011 01:08:27 +0000

Sam already wrote about the g+hangout that he hosted last Saturday and I wanted to jot down some of my own thoughts on the experience. I thoroughly enjoyed the whole thing and I’ve been looking for good imagery to describe …

The Secret Santa Problem

Micheal Pawliuk — Tue, 20 Dec 2011 22:45:51 +0000

Happy holidays everyone! As Christmas approaches so do the Christmas related problems. I’m not talking about the long lines at stores or the busy days filled with errands, I’m talking about Christmas math problems. Here is one I learned at my department holiday party from Eric Hart and Jeremy Voltz.

This whole post is going to be directed at a general audience.

Secret Santa Problem (simplified). An office needs to determine how to set up a secret santa gift exchange, but they have lost all of their dice and paper! How can the each person in the office have exactly one person for whom they are buying a gift and also do not know who is buying them a gift?

Here we allow some of the employees to have private conversations if they wish.

Attempt 1: The obvious first thing to do (that doesn’t work) is to have one person tell everyone whose gift they are buying. You can tell right away why this won’t work: ~~It is too much work for that one person~~ the designator will have to designate someone to buy a gift for the designator!

Attempt 1.a: Have the boss tell everyone what they should do. Well… I don’t think the boss is going to like this idea. We should really try to find an internal solution. That is, let’s try to find a solution that does not use anything external (like random number generators, extra people, secret santa consultants, etc.)

What else obviously doesn’t work?

Attempt 2: Everyone could just pick another employee privately. This won’t work because Larry from accounting might get 100 gifts. There is no way to guarantee that this method will ensure that no employee gets more than one gift.

Attempt 3: Have everyone sit in a circle and buy a gift for the person to the left of them. Think about why this doesn’t work. This attempt is very similar to one of the previous solutions (which one?).

In general, finding a bunch of methods that don’t work is a good way to start tackling a problem.

“How often have I said to you that when you have eliminated the impossible, whatever remains, however improbable, must be the truth?” -Sherlock Holmes

Hopefully, if we can find the different extremes of wrong methods we can guess at a middle ground (or “interpolate”). If the extremes are sufficiently different (or “orthogonal”) then we would like to blend these methods together to get a correct solution. This is all very vague, but the more problem solving you do the more helpful this strategy becomes.

Let us look at the pros and cons of the attempts we have made:

Attempt #	Pros	Cons
1	Easy to describe, only one “mistake”, lots of secrecy	Asymmetric
1.a	Easy to describe, completely secret	Asymmetric, external solution
2	Easy to describe, symmetric, lots of secrecy	Many mistakes possible
3	Easy to describe, symmetric	Too much communal knowledge

What do we notice here? All of these attempts are easy to describe and it seems like we are balancing “secrecy” and “symmetry”. It looks like we have come up with some attempts that hit the extremes of high/low secrecy and high/low symmetry. Maybe we can blend these attempts to get something that is mostly secret and mostly symmetric.

Here is your hint: What if instead of employees we have couples of employees? (Think about this a bit before reading on. I’ll wait.)

…

Any ideas? Well, simplifying a bit let’s look at some small numbers. The problem clearly doesn’t make sense with only one or two people. It also doesn’t work for three people. (Can you see how there just aren’t enough options?) So, let’s try this with four people. Using the hint, let’s say the four people are me, my wife Janet, Sam Coskey and his wife Kathy.

What is the “natural” way of setting up this secret santa?

Attempt 4: Well Janet and I will buy gifts for Sam and Kathy, but we will not tell them for whom specifically we are buying our individual gifts. Sam and Kathy will do a similar thing but buying gifts for me and Janet. Here I know that Sam or Kathy is buying me a gift, but I don’t know for sure which person it is. In this method we have sacrificed a bit of the secrecy (I know that Janet is not buying my gift) and retained a type of symmetry from attempt 2.

For interest’s sake, here are the pictures that I thought of to go with the different attempts. (Please excuse the quality):

The next question is the full secret Santa Problem:

Secret Santa Problem. An office needs to determine how to set up a secret santa gift exchange, but they have lost all of their dice and paper! How can the each person in the office have exactly one person for whom they are buying a gift and also do not know who is buying them a gift? Moreover, any given person has no information about who could be their secret santa.

I’m not going to attempt this one, instead I’m going to end with a similar question that Sam Coskey gave me over supper:

Sam’s 2-person “secret santa”. Is it possible for two people to each choose a natural number so that both numbers are exactly 1 apart and neither person knows who has the larger of the two numbers?

And a cute generalization:

Father/Son “secret santa”. Is it possible for two players to each choose a human being so that the two humans are father and son, but neither player knows who has the father and who has the son?

I will tackle these problems next week. Until then, happy holidays!

The logic of Reverse Mathematics

Carl Mummert — Tue, 20 Dec 2011 02:46:11 +0000

This post is about a research idea I have been thinking about which is quite different from my usual research. It’s an example of a project with an “old fashioned” feel to it, as if it could have been studied … Continue reading →

The recent publishing debate — the IMU’s blog continues

Peter Krautzberger — Fri, 16 Dec 2011 20:37:44 +0000

My last post has seen quite a lot of visitors (my stats look miserable now that there was such a spike…). But hardly anybody left a comment (thanks to those who did — much appreciated!). Did people find it irrelevant, …

Why write papers?

François G. Dorais — Tue, 13 Dec 2011 23:09:01 +0000

Here is what Dror Bar-Natan had to say about that on MathOverflow: Papers are written so that their author(s) can forget their content and move on to other things. Therefore when you write you should be very careful to put in enough of the big picture and enough of the details so you’d be able [...]

Czech Winter School

KT — Mon, 12 Dec 2011 14:20:57 +0000

I just registered for the Winterschool in Abstract Analysis section Set Theory and Topology which takes place 28 Jan. – 4 Feb. 2012. Registration deadline is coming up at the end of the month! I’m really looking forward to this.

The recent publishing debate — The economic power of publishers

Peter Krautzberger — Mon, 12 Dec 2011 05:20:25 +0000

I have been trying to find the time to continue my posts on the publishing debate, discussing the other posts from the original timeline. Suffice it to say, I’ve not yet given up getting around to it… But there were
…

Reading the Dictionary

Micheal Pawliuk — Tue, 29 Nov 2011 02:08:23 +0000

I have a confession to make: I am a bibliophile. Reading, owning, perusing, lending, alphabetizing and buying books are all things that make me happy. High on my list are hardcover graphic novels and quality dictionaries. One of the skills you learn quickly while reading a dictionary (so I hear) is how to look up words. Of course the words in a dictionary are laid out in a very orderly fashion; first the ‘A’s then the ‘B’s, etc.. This order turns out to be a useful example of an interesting linear order.

Example: Consider $\{a,b,c\}\times\{a,b\}$ with the dictionary ordering. We get $aa < ab < ba < bb < ca < cb$.

In general to get a dictionary ordering on $A\times B$ out of two linear orders $A,B$ we do the following:

Compare first elements. If they are the different, use the ordering on A.
If the first coordinates are different, compare the second coordinates. If the second coordinates are different, use the ordering on $B$. If the second coordinates are the same, the elements you are comparing are the same (as they have the same first and second coordinates).

You can extend this process if you want and compare third, fourth or fifth coordinates if you start with three, four or five linear orders. Of course this is just saying something you already know; I don’t need to tell you how to figure out whether ‘oscillate’ comes before ‘ossifrage‘.

Example: Now my fellow sesquipedalians might be interested in the following linear order: Let $D = \{*, a,b,c, \ldots, z\}$ where $* < a < b < \ldots < z$ and $*$ stands for a blank space. Now consider $D^{189819}$ with the dictionary ordering. This will contain every English word both technical and non-technical. Granted it will also contain silly non-words like: “this*word*asserts*that*it*is*a*silly*word”.

Now, down to business. Recall that any linear order can be imbued with a topology called the linear order topology.

Linear Order Topology. Let $(L, <)$ be a linear order. The linear order topology is given by the subbasic open sets $(-\infty, a) := \{x \in L : x <a\}$ and $(b, \infty) := \{x \in L : b <x\}$.

A basic open set.

Here the use of the symbol $\infty$ is purely for notational convenience. In most cases we will be able to say that the intervals $(a,b)$ form a basis for the linear order topology.

Example: The usual topology on $\R$ is given by the linear order topology as every open ball is an open interval and every open interval is a union of open balls. Technically, for the masochists: $$(\forall a,b \in \R \cup \{\infty,-\infty\})(\forall x \in (a,b))(\exists \epsilon > 0)(\exists y \in \R)[x \in B_\epsilon (y) \sse (a,b)]$$

Non-example: The Sorgenfrey line $\R_l$ is not given by an order topology. This can be seen by seeing that the order topology on $\R$ is the usual topology, which is not the Sorgenfrey topology.

Example: Looking at $\{0,1\} \times \R$ with the dictionary ordering, we can give it the order topology. What does this look like? Well it looks like two copies of $\R$ sitting next to each other. Everything in the $\{0\}\times\R$ interval is less than everything in the $\{1\}\times\R$ interval. Or, if you prefer, it is the same as $(3,4)\cup(5,6)$.

Now what I am really concerned with are the following two spaces, each with the dictionary ordering: $[0,1]\times[0,1]$ and $(0,1)\times(0,1)$. On first blush they seem like they should have some similar properties.
$(0,1)\times(0,1)$. The first thing we notice is that this space is not separable, Lindelöf or second countable, all for (basically) the same reason: The set $\{(x,\frac{1}{2}): x \in (0,1)\}$ is an uncountable, closed discrete set. However, it is first-countable, for around any point there is the obvious countable local base in the second coordinate. Clearly it is not, compact or limit-point compact. As for separation properties, it is obviously Hausdorff and (non-obviously) normal. One way to see this is that a set $C \sse (0,1)\times(0,1)$ is closed iff when I take everything in $C$ with a given fixed first coordinate, it is closed in that coordinate.
Showing metrizabilty of this space is an (easy) exercise in Munkres. I thought about it for a long time. It turns out that this space is really just the product of two metric spaces: $(0,1)$usual and $(0,1)$discrete. Looking at it this way is somewhat illuminating: the different coordinates are actually far apart, and the discrete metric captures that.
So to sum up:

	$(0,1)\times(0,1)$
T2	YES!
T4	YES!
Compact	NO.
Limit Point Compact	NO.
Lindelöf	NO.
Separable	NO.
Second Countable	NO.
First Countable	YES!
Metrizable	YES!

$[0,1]\times[0,1]$: So think about this space for a bit. It looks awfully similar to the last space. Looks like this one is also just the product of two metric spaces. Well let’s see. For the same reasons as the last space, $[0,1]\times[0,1]$ is not separable, not second countable, and is Hausdorff. Is this space first countable? Yes, but it takes a bit of thinking to see why the endpoints have a countable local base. For example, what happens at the point $(\frac{1}{2}, 1)$?

Now the best part is that $[0,1]\times[0,1]$ is compact. Hold on a second before you say: “Clearly, Mike”, because you can’t use Tychonoff’s Theorem here. This is not $[0,1]\times[0,1]$ with the product topology. However, we can prove that it is compact by creeping along! The only technicality is showing that, as a linear order, $[0,1]\times[0,1]$ is complete but this happens mostly by inheriting completeness from $\R$.
So now we know that $[0,1]\times[0,1]$ cannot be metrizable (!) as compact and metrizability give separability, which this space does not have. Also, this space must be limit point compact; doesn’t that seem counterintuitive from our experience with $(0,1)\times(0,1)$! Also, we get normality since the space is compact and Hausdorff.
What went wrong with our argument about this space being the product of two metric spaces? Well, we didn’t actually check that the topologies agree, and in fact they don’t. In the metric topology any individual coordinate is both open and closed. In the order topology though, individual coordinates cannot be open, because open sets that contain an endpoint also contain a lot of other coordinates.

So to summarize:

	$(0,1)\times(0,1)$	$[0,1]\times[0,1]$
T2	YES!	YES!
T4	YES!	YES!
Compact	NO.	YES!
Limit Point Compact	NO.	YES!
Lindelöf	NO.	YES!
Separable	NO.	NO.
Second Countable	NO.	NO.
First Countable	YES!	YES!
Metrizable	YES!	NO.

I think that these two spaces are fairly simple, possess nice properties, but seem to be relatively unknown.

The recent publishing debate — Nisan’s posts

Peter Krautzberger — Sun, 27 Nov 2011 19:19:38 +0000

The original post was getting longer and longer so I split it up.

To refresh your memory,

Nov 2 Algorithmic Game-Theory/Economics // Noam Nisan: The problem with journals
Nov 3 Algorithmic Game-Theory/Economics // Noam Nisan: The good things

XKCD on the Axiom of Choice…

François G. Dorais — Sun, 27 Nov 2011 16:05:50 +0000

Arguably one of the best XKCD comics ever… The alternate text for this comic is a modern lumberjack’s proof of Zermelo’s Theorem. Proof of Zermelo’s well-ordering theorem given the Axiom of Choice: 1: Take S to be any set. 2: When I reach step three, if S hasn’t managed to find a well-ordering relation for [...]

The recent publishing debate — a timeline

Peter Krautzberger — Sun, 27 Nov 2011 00:44:26 +0000

In the last 3 weeks I have written a couple of drafts about the debate that finally hit the mathematical blogosphere through Tim Gowers’s blog (I don’t know how much he is aware of similar, ongoing discussions in the scientific …

The single greatest influence

Peter Krautzberger — Sun, 20 Nov 2011 17:47:30 +0000

Note: this is an old draft that I finally finished; there’s a line to indicate the major break between the original half and the additional part (not that the first part wasn’t rewritten a little, too).

Yesterday, when hearing the …

Moving to WordPress

Carl Mummert — Sun, 20 Nov 2011 02:29:46 +0000

For a while I have been thinking about migrating to a content management system for my web page. Previously I had a script I wrote that was essentially a simply wiki to let me edit pages from any web browser … Continue reading →

Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)

Peter Krautzberger — Wed, 16 Nov 2011 03:58:27 +0000

Because of a power outage at the department my talk announced for October 29th was postponed by a week.

Idempotent Ultrafilters: An Introduction (University of Michigan Logic Seminar 2011-11-08) from Peter Krautzberger on Vimeo.

Here are transcripts of my …

Inner models with large cardinal features usually obtained by forcing

Victoria Gitman — Thu, 10 Nov 2011 16:29:36 +0000

A. Apter, V. Gitman, and J. D. Hamkins, “Inner models with large cardinal features usually obtained by forcing,” Archive for Mathematical Logic, vol. 51, pp. 257-283, 2012. (10.1007/s00153-011-0264-5)

PDF Journal Citation arχiv

@article {apterhamkinsgitman:inner,
author = {Apter, Arthur and Gitman, Victoria and Hamkins, Joel David},
affiliation = {Mathematics, The Graduate Center of the City University of New York, 365 Fifth Avenue, New York, NY 10016, USA},
title = {Inner models with large cardinal features usually obtained by forcing},
journal = {Archive for Mathematical Logic},
publisher = {Springer Berlin / Heidelberg},
issn = {0933-5846},
keyword = {Mathematics and Statistics},
pages = {257--283},
volume = {51},
issue = {3},
PDF={http://boolesrings.org/victoriagitman/files/2011/08/innermodels.pdf},
eprint = {1111.0856},
doi = {10.1007/s00153-011-0264-5},
note = {10.1007/s00153-011-0264-5},
year = {2012}
}

This is joint work with Arthur Apter and Joel David Hamkins.

In this article, we provide techniques for dealing with questions asking of a particular set-theoretic assertion known to be forceable over a universe with large cardinals, whether it must hold already in an inner model whenever such large cardinals exist. The questions therefore concern what we describe as the internal consistency strength of the relevant assertions, a concept we presently explain. Following ideas of Sy Friedman [1], let us say that an assertion $\varphi$ is internally consistent if it holds in an inner model, that is, if there is a transitive class model of ${\rm ZFC}$, containing all the ordinals, in which $\varphi$ is true. In this general form, an assertion of internal consistency is a second-order assertion, expressible in ${\rm GBC}$ set theory; nevertheless, it turns out that many interesting affirmative instances of internal consistency are expressible in the first-order language of set theory, when the relevant inner model is a definable class, and as a result much of the analysis of internal consistency can be carried out in first-order ${\rm ZFC}$. One may measure what we refer to as the internal consistency strength of an assertion $\varphi$ by the hypothesis necessary to prove that $\varphi$ holds in an inner model. Specifically, we say that the internal consistency strength of $\varphi$ is bounded above by a large cardinal or other hypothesis $\psi$, if we can prove from ${\rm ZFC}+\psi$ that there is an inner model of $\varphi$; in other words, if we can argue from the truth of $\psi$ to the existence of an inner model of $\varphi$. Two statements are internally-equiconsistent if each of them proves the existence of an inner model of the other. It follows that the internal consistency strength of an assertion is at least as great as the ordinary consistency strength of that assertion, and the interesting phenomenon here is that internal consistency strength can sometimes exceed ordinary consistency strength. For example, although the hypothesis $\varphi$ asserting “there is a measurable cardinal and ${\rm CH}$ fails” is equiconsistent with a measurable cardinal, because it is easily forced over any model with a measurable cardinal, nevertheless the internal consistency strength of $\varphi$, assuming consistency, is strictly larger than a measurable cardinal, because there are models having a measurable cardinal in which there is no inner model satisfying $\varphi$. For example, in the canonical model $L[\mu]$ for a single measurable cardinal, every inner model with a measurable cardinal contains an iterate of $L[\mu]$ and therefore agrees that ${\rm CH}$ holds. So one needs more than just a measurable cardinal in order to ensure that there is an inner model with a measurable cardinal in which ${\rm CH}$ fails.

We construct a variety of inner models exhibiting features usually obtained by forcing over universes with large cardinals. For example, if there is a supercompact cardinal, then there is an inner model with a Laver indestructible supercompact cardinal. If there is a supercompact cardinal, then there is an inner model with a supercompact cardinal $\kappa$ for which $2^\kappa=\kappa^+$, another for which $2^\kappa=\kappa^{++}$ and another in which the least strongly compact cardinal is supercompact. If there is a strongly compact cardinal, then there is an inner model with a strongly compact cardinal, for which the measurable cardinals are bounded below it and another inner model $W$ with a strongly compact cardinal $\kappa$, such that $H_{\kappa^+}^V\subseteq{\rm HOD}^W$. Similar facts hold for supercompact, measurable and strongly Ramsey cardinals. If a cardinal is supercompact up to a weakly iterable cardinal, then there is an inner model of the Proper Forcing Axiom and another inner model with a supercompact cardinal in which ${\rm GCH}+V={\rm HOD}$ holds. Under the same hypothesis, there is an inner model with level by level equivalence between strong compactness and supercompactness, and indeed, another in which there is level by level inequivalence between strong compactness and supercompactness. If a cardinal is strongly compact up to a weakly iterable cardinal, then there is an inner model in which the least measurable cardinal is strongly compact. If there is a weakly iterable limit $\delta$ of ${<}\delta$-supercompact cardinals, then there is an inner model with a proper class of Laver-indestructible supercompact cardinals. We describe three general proof methods, which can be used to prove many similar results.

[1] S. Friedman, “Internal consistency and the inner model hypothesis,” Bull. Symbolic Logic, vol. 12, iss. 4, pp. 591-600, 2006.
[Bibtex]

@article {Friedman2006:InternalConsistencyAndIMH,
AUTHOR = {Friedman, Sy-David},
TITLE = {Internal consistency and the inner model hypothesis},
JOURNAL = {Bull. Symbolic Logic},
FJOURNAL = {Bulletin of Symbolic Logic},
VOLUME = {12},
YEAR = {2006},
NUMBER = {4},
PAGES = {591--600},
ISSN = {1079-8986},
MRCLASS = {03E35 (03E45 03E55)},
MRNUMBER = {2283091 (2007j:03065)},
MRREVIEWER = {Qi Feng},
URL = {http://projecteuclid.org/getRecord?id=euclid.bsl/1164056808},
}

On a theorem of Mycielski and Taylor

François G. Dorais — Sun, 06 Nov 2011 19:28:03 +0000

$\DeclareMathOperator{\cov}{cov}\newcommand{\meager}{\mathcal{M}}$ In 1964, Jan Mycielski [1] proved a wonderful theorem about independent sets in Polish spaces. He showed that if $X$ is an uncountable Polish space and $R_n$ is a meager subset of $X^n$ for each $n \geq 1$, then there is a perfect set $Z \subseteq X$ such that $(z_1,\dots,z_n) \notin R_n$ whenever $z_1,\dots,z_n$ [...]

Uncountable perfect graphs

François G. Dorais — Sun, 06 Nov 2011 15:03:58 +0000

I have been fascinated with perfect graphs for many years. These graphs are very intensely studied in finite combinatorics and new wonderful properties of finite perfect graphs are discovered on a regular basis. However, it is still unclear whether or how these wonderful properties might extend to the countable and uncountable cases. There are several [...]

Two perspectives on Halloween

Victoria Gitman — Fri, 04 Nov 2011 16:04:45 +0000

I went trick-or-treating for the first time this Halloween, twenty years since coming to the United States. Incidentally, according to Wikipedia, Halloween has its origins in the celtic festival Samhain, a harvest celebration and a time to honor the dead. Way back in the spring, I decided, after seeing the previews for the new movie, to be the beautiful brave red riding hood. I guilt tripped my mom into making the costume after much convincing that her motherly duties still applied at this age. Mom was quite crafty in her former communist life out of necessity but since coming to the United States had decided that there is little point in making things by hand when they are made much better and cheaper in China. She kept searching Amazon in hopes of avoiding the said duties and finally caved in when just the right costume appeared at the price of $100 plus shipping. The weekend’s freak snow storm along with a dire warning from Metuchen’s mayor about live wires and falling branches looked to put end to my first Halloween adventure. Luckily, by Monday afternoon the sun was shining and the left over patches of snow only added to the magical atmosphere of fairy creatures and sweets. Metuchen’s kids (and supervising adults) were out in full force drawn into the world of fantasy come to life, periodically switching the roles of collecting and giving out sweets. I even started to fret that my Costo super-sized candy bag would run out. I trick or treated in the official role of assistant chaperon to the daughter of a friend and gallantly donated the sweet complements to my costume to her burgeoning bag. After a glorious hour of candy collecting and socializing with all manner of critters big and small, we found our way back to my friend’s house. There her husband, oblivious, to all din and hijinks of the outside world, was painstakingly reproducing a haunting Asian painting of a mountainous landscape in which two specks of human figures trudged up a slope dominated by enormous trees. We doubted that he heard the many merry pleas for candy coming from the kids at his front door. I called him a Grinch and wondered whether he was secretly hatching a plan to steal Christmas. He responded that fully giving himself up to painting in complete disregard of the nonsensical spectacle outside was “life affirming”. It shall remain one of those mysteries of the universe which one of us truly lived up to the spirit of Samhain.

Red Riding Hood

Forcing and gaps in $2^\omega$

Victoria Gitman — Thu, 03 Nov 2011 18:51:47 +0000

This is a talk at the CUNY Set Theory Seminar, December 2nd, 2011.

For $a,b\in 2^\omega$, we say that $a$ is eventually dominated by $b$, denoted by $a\leq^*b$, if $a(n)\leq b(n)$ for all but finitely many $n$. Let $\mathcal A=\langle a_\alpha\mid \alpha<\kappa\rangle$ and $\mathcal B=\langle b_\beta\mid \beta<\lambda\rangle$, where $\kappa$ and $\lambda$ are infinite regular cardinals, be a pair of sequences in $2^\omega$. The pair $(\mathcal A,\mathcal B)$ is called a $(\kappa,\lambda)$-pregap if $a_{\alpha_1}\leq^*a_{\alpha_2}\leq^*b_{\beta_2}\leq^* b_{\beta_1}$ for all $\alpha_1<\alpha_2<\kappa$ and $\beta_1<\beta_2<\lambda$. That is, we have:

$a_0\leq^* a_1 \leq^*\cdots\leq^* a_\alpha \leq^*\cdots \leq^* b_\beta\leq^* \cdots \leq^* b_1 \leq^* b_0$

We say that a set $c\in 2^\omega$ separates the pregap $(\mathcal A,\mathcal B)$ if $a_\alpha\leq^* c\leq^* b_\beta$ for all $\alpha<\kappa$ and $\beta<\lambda$. That is, we have:

$a_0\leq^* a_1 \leq^*\cdots\leq^* a_\alpha \leq^*\cdots\leq^* c\leq^*\cdots \leq^* b_\beta\leq^* \cdots \leq^* b_1 \leq^* b_0$>

If there is no such set $c$, then we say that the pregap $(\mathcal A,\mathcal B)$ is a $(\kappa,\lambda$)-gap. Much of the literature on gaps also studies gaps in $\omega^\omega$ under the eventual domination ordering. However, it is not difficult to see that for infinite regular cardinals $\kappa$ and $\lambda$, there is a $(\kappa,\lambda)$-gap in $\omega^\omega$ if and only if there is one in $2^\omega$. Thus, we can concentrate on gaps in $2^\omega$ without loss of generality. In what follows we tacitly associate elements of $2^\omega$ with subsets of $\omega$.

(Hadamard, 1894) There are no $(\omega,\omega)$-gaps. Consider a pregap $(\mathcal A,\mathcal B)$, where $\mathcal A=\langle a_n\mid n<\omega\rangle$ and $\mathcal B=\langle b_m\mid m<\omega\rangle$. Let $\overline b_m$ denote the complement of $b_m$ and define $c_n=a_n\setminus (\bigcup_{m\leq n} \overline b_m)$. It is now easy to see that $c=\bigcup_{n<\omega} c_n$ separates $(\mathcal A,\mathcal B)$. (Hausdorff, 1909) There is an $(\omega_1,\omega_1)$-gap.

For a proof see [1] (Section 29).

Here, we focus on the interaction between $(\omega_1,\omega_1)$-gaps and forcing. In particular, we are interested here in the following questions:
Can we create a generic $(\omega_1,\omega_1)$-gap by $\omega_1$-preserving forcing?

Let us call an $(\omega_1,\omega_1)$-gap destructible, if there is an $\omega_1$-preserving forcing which adds a set separating it. Note that every $(\omega_1,\omega_1)$-gap is trivially destructible, if we remove the requirement that the forcing is $\omega_1$-preserving, by collapsing $\omega_1$ to $\omega$. We call an $(\omega_1,\omega_1)$-gap indestructible if it is not destructible. Kunen showed (1976) that Hausdorff's gap is indestructible.

Are there destructible $(\omega_1,\omega_1)$-gaps?

Can we force to make an $(\omega_1,\omega_1)$-gap indestructible?

The material in this talk draws mainly on Teruyuki Yorioka’s thesis [2] and Marion Scheepers’ survey paper [3]. For full notes see PDF.

[1] T. Jech, Set theory, Berlin: Springer-Verlag, 2003.
[Bibtex]

@book {jech:settheory,
AUTHOR = {Jech, Thomas},
TITLE = {Set theory},
SERIES = {Springer Monographs in Mathematics},
NOTE = {The third millennium edition, revised and expanded},
PUBLISHER = {Springer-Verlag},
ADDRESS = {Berlin},
YEAR = {2003},
PAGES = {xiv+769},
ISBN = {3-540-44085-2},
MRCLASS = {03Exx (03-01 03-02)},
MRNUMBER = {1940513 (2004g:03071)},
MRREVIEWER = {Eva Coplakova},
}

[2] T. Yorioka, “Some results on gaps in $P(\omega)/{\rm fin}$,” PhD Thesis, 2004.
[Bibtex]

@PHDTHESIS{yorioka:thesis,
AUTHOR={Teruyuki Yorioka},
TITLE={Some results on gaps in {$P(\omega)/{\rm fin}$}},
SCHOOL={Kobe University},
YEAR ={2004}}

[3] M. Scheepers, “Gaps in $\omega^\omega$.” Ramat Gan: Bar-Ilan Univ., 1993, vol. 6, pp. 439-561.
[Bibtex]

@incollection {scheepers:gaps,
AUTHOR = {Scheepers, Marion},
TITLE = {Gaps in {$\omega^\omega$}},
BOOKTITLE = {Set theory of the reals ({R}amat {G}an, 1991)},
SERIES = {Israel Math. Conf. Proc.},
VOLUME = {6},
PAGES = {439--561},
PUBLISHER = {Bar-Ilan Univ.},
ADDRESS = {Ramat Gan},
YEAR = {1993},
MRCLASS = {03E05 (03E35 03E50 03E65 04-02 06A07)},
MRNUMBER = {1234288 (95a:03061)},
MRREVIEWER = {Pierre Matet},
}

Classically valid theorems of intuitionistic analysis

François G. Dorais — Sun, 30 Oct 2011 18:14:30 +0000

It is well-known that proofs in intuitionistic logic are more constructive than proofs in classical logic. Indeed, this is what the (informal) Brouwer–Heyting–Kolmogorov (BHK) interpretation leads to believe. Thus, intuitionistic proofs tend to have more information content than classical proofs, a fact that is well exploited in proof mining. This suggests that looking at intuitionistic [...]

Oracle forcing part II

KT — Thu, 27 Oct 2011 13:40:04 +0000

Putting oracle forcing into context

$\sigma$-closed
$\Downarrow$
uberoracle-proper	$\Rightarrow$	strong $\bar{M}$-proper	$\Longrightarrow$ $\Rightarrow$ $\bar{M}$-proper			proper
$\Uparrow$		$\Uparrow$		$\Uparrow$		$\Uparrow$
uberoracle-cc	$\Rightarrow$	strong $\bar{M}$-cc	$\Rightarrow$	$\bar{M}$-cc	$\Rightarrow$	ccc
						$\Uparrow$
						$\sigma$-centred

Note: All arrows are strict. The only arrow which could possibly be added is $\sigma$-centred $\Rightarrow \bar{M}$-proper; this is unknown. The picture is supposed to show that strong oracle-proper implies proper, but oracle-proper does not imply proper.

Proofs of these implications and counterexamples to show strictness are given below (at least the ones that are not obvious, well-known or mentioned in the previous post), working our way from left to right in the picture.

Claim Any $\sigma$-closed forcing is $\bar{M}$-proper for any oracle $\bar{M}$.

Proof Starting with a condition $p$ and $\delta \in \omega_1$, basically use the closure of the forcing to extend $p$ to a $(\delta, M_\delta)$-generic condition.
$\Box$

To see that uberoracle-cc implies uberoracle-proper and that strong oracle-cc implies strong oracle-proper, one simply uses the fact that if a forcing is $\bar{M}$-cc for a specific oracle $\bar{M}$ then it is $\bar{M}$-proper for the same oracle (see Abraham’s notes for a proof of this latter fact).

The following claim (proof due to Martin is forthcoming) implies both that the notion of uberoracle-proper is strictly stronger than oracle-proper and that uberoracle-cc is stronger than oracle-cc.

Claim:

For any oracle $\bar{M}$ there is a Suslin tree $T$ such that $T$ is $\bar{M}$-cc.
For any Suslin tree $T$ and any oracle $\bar{M}$ there is $\bar{M}^\prime \trianglerighteq \bar{M}$ such that $T$ is not $\bar{M}^\prime$-proper.

The next claim can be found in Shelah 100.

Claim For any strong oracle $\bar{M}$, if $P$ is strong $\bar{M}$-proper and $|P| \leq \aleph_1$ then $P$ is proper.

Proof Assume $P = \omega_1$. Fix $p \in P$. Let $N$ be a countable elementary submodel of some large $H(\xi)$ such that $\bar{M}, P \in N$. Let $\delta^* = N \cap \omega_1$.

Let $S$ be the set of all $\delta < \omega_1$ such that there exists $q \leq p$ where $q$ is $(\delta, M_\delta)$-generic which is club as $P$ is $\bar{M}$-proper and $\bar{M}$ is a strong oracle. Therefore we may assume that that $\delta^* \in S$.

Now let $E \in N$ such that $E \subseteq P$ and dense in $P$. The set $C$ of $\delta \in \omega_1$ such that $E \cap P \upharpoonright \delta$ is dense in $P\upharpoonright \delta$ is club and $C \in N$ so $\delta^* \in C$. Also $E \cap P \upharpoonright \delta^* \in N$. We want to see that $E \cap P\upharpoonright \delta^* \in M_{\delta^*}$.

The set
$$\{N : N \cap [\omega_1]^{\aleph_0} \subseteq M_{(N \cap \omega_1)}\}$$
is club in $[H(\xi)]^{\aleph_0}$.
Why? Given $N_i : i < \omega$ in this set, we have
$$(\bigcup N_i) \cap [\omega_1]^{\aleph_0} \subseteq \bigcup_i M_{(N_i \cap \omega_1)} \subseteq M_{(\bigcup N_i \cap \omega_1)}$$
so it is closed. Thus the $N$ fixed above is a member of this set, which gives
$$E \cap P\upharpoonright \delta^* \in M_{\delta^*}.$$

Now by $\bar{M}$-properness we have $E \cap P\upharpoonright \delta^* = E \cap N$ is predense in $P$ below $q$.

$\Box$

This next one though, proved by Martin, comes as somewhat a surprise and makes us think twice about bothering at all with the weak definition of oracle-proper.

Claim: There is a $P$ which is $\bar{M}$-proper and $|P| = \aleph_1$ but $P$ is not proper.

Proof Let $S$ be a stationary, co-stationary subset of $\omega_1$ and let $P$ be the forcing which collapses $\omega_1 \setminus S$. That is, conditions are continuous functions $f : \alpha \rightarrow \alpha$ where $\alpha$ is a successor ordinal and $f(\beta) \in S$ for all $\beta < \alpha$. This forcing is not proper, as models $N$ such that $N \cap \omega_1 \not\in S$ do not have $(N,P)$-generic conditions.

Let $\bar{M}$ be defined as $\{M_\delta : \delta \in S\}$. Given $\delta \in S$ and $p \in M_\delta$ we will find $q \leq p$ which is $(\delta, M_\delta)$-generic. Denote by $P \upharpoonright \delta = \{f \in P : f \subset \delta \times \delta\}$. Enumerate by $\langle A_n : n < \omega\rangle$ the set of $A \subseteq P\upharpoonright \delta$ such that $D \in M_\delta$ and are antichains in $P\upharpoonright \delta$. We may extend $p$ in $\omega$-steps such that each $p_n$ forces that the generic intersects $A_n$ at a point $a_n$. Then $q = \bigcup p_n \cup \{(\delta, \delta)\}$ is a condition in $P$ and forces that for all $n < \omega$ the $P$-generic filter meets $A_n$ at $a_n$ (i.e. is non-empty).
$\Box$

Finally we show that the oracle-cc to oracle-proper implications are strict.

Claim: There is a forcing which is uberoracle-proper but neither $\omega$-proper nor ccc.

Proof Let conditions in $P$ be finite partial functions $p : \omega_1 \rightarrow \omega_1$ which are weakly increasing. This is proper, but not Axiom A, see Jech Ch. 31 exercises.

To see that this forcing is oracle proper for any oracle, let $p \in P$ and $\delta$ be such that $P \upharpoonright \delta = \{p \upharpoonright (\delta \times \delta) : p \in P\}$ (happens on a club). Then for $M_\delta$ we let $q = p \cup \{\delta, \delta\}$ which is $(\delta, M_\delta)$-generic.
$\Box$

On computing complex square roots

François G. Dorais — Sat, 22 Oct 2011 16:15:34 +0000

What could possibly be interesting about complex square roots? That’s what I thought until very recently when I realized that complex square roots are significantly more difficult to compute than real square roots. Let me explain why by putting you in a context where you need to teach students how to compute complex square roots… [...]

Indestructibility for Ramsey and Ramsey-like cardinals

Victoria Gitman — Fri, 21 Oct 2011 19:34:58 +0000

V. Gitman and T. A. Johnstone, “Indestructibility for Ramsey and Ramsey-like cardinals.” (In preparation)

Citation

@ARTICLE{gitman:ramseyindes,
AUTHOR= {Victoria Gitman and Thomas A. Johnstone},
TITLE= {Indestructibility for {R}amsey and {R}amsey-like cardinals},
NOTE= {In preparation}}

This is joint work with Thomas Johnstone.

The technique of forcing, introduced by Paul Cohen in the early 1960′s, proved to be one of the most powerful tools for producing independence and consistency results in set theory. Starting with a partial order (the notion of forcing) in a given universe of set theory (the ground model) a new universe (the generic extension), is constructed as the closure of the ground model and a generic filter for that partial order. The generic extension is always a ${\rm ZFC}$-universe and a cleverly chosen partial order will force it to satisfy set theoretic properties of interest that may have been false in the ground model. If the goal is to show the consistency of a set theoretic property with a large cardinal, an immediate complication is that the large cardinal may be destroyed by the partial order that forces the property in the generic extension, by, for instance, collapsing it to $\omega_1$. The decades following the introduction of forcing saw the development of a toolkit of techniques for showing that a forcing notion preserves
the large cardinal outright or does so over a ground model prepared through forcing over the original universe. A large cardinal that is preserved by a forcing notion is said to be indestructible by it. A seminal result in the study of indestructibility from the 1960′s, the Lévy-Solovay theorem [1], showed that measurable cardinals are indestructible by small forcing (small in size relative to the cardinal) and consequently cannot decide such properties as the continuum hypothesis, existence of Suslin trees, or the $\Diamond$-principle, which can be made to hold or fail with small forcing relative to an inaccessible cardinal. Most large
cardinal notions are now known to be indestructible by small forcing. Silver pioneered techniques for preparing a ground model using long forcing iterations (see [2] for an exposition) which show that a measurable cardinal or a weakly compact cardinal becomes indestructible by ${\rm Add}(\kappa,1)$ in some forcing extension. Since a weakly compact cardinal $\kappa$ indestructible by ${\rm Add}(\kappa,1)$ is automatically indestructible by ${\rm Add}(\kappa,1)$ for all $\theta$, it follows every weakly compact $\kappa$ becomes indestructible by ${\rm Add}(\kappa,\theta)$ for all $\theta$ in some forcing extension. Contrastingly, every measurable cardinal $\kappa$ cannot cannot become indestructible by ${\rm Add}(\kappa,\kappa^{++})$ in some forcing extension since the consistency strength of a measurable cardinal at which the ${\rm GCH}$ fails is greater than a measurable cardinal [3], thus providing an illustration of a negative indestructibility result. General indestructibility results covering a wide class of forcing notions are known only for supercompact cardinals and strongly unfoldable cardinals. Laver showed in [4] that a supercompact cardinal becomes indestructible by all ${<}\kappa$-directed closed forcing in a forcing extension, and [5] showed that a strongly unfoldable cardinal becomes indestructible by all ${<}\kappa$-closed $\kappa^+$-preserving forcing in a forcing extension.

Very little was previously known about the indestructibility properties of Ramsey cardinals and the few existing results lacked a uniform set of techniques. Using their combinatorial definition, it was shown that Ramsey cardinals are indestructible by small forcing [6] (Section 10) and Jensen in [7] hinted at a proof that Ramsey cardinals are indestructible by a forcing which yields the ${\rm GCH}$ in the forcing extension. Finally, Welch showed in [8], using the characterization of Ramsey cardinals in terms of the existence of good sets of indiscernibles, that they are indestructible by the forcing to code the universe into a real.

Techniques for showing indestructibility chiefly rely on the characterization of a large cardinal in terms of the existence of elementary embeddings. The basic idea is to argue that the large cardinal is preserved by lifting the ground model embedding to the forcing extension. The relationship between the existence of elementary embeddings and large cardinals is one of the central unifying themes in the theory of large cardinals. Measurable cardinals and most stronger large cardinals are characterized by the existence of elementary embeddings $j:V\to M$ from the universe into an inner model with that cardinal as the critical point. Smaller large cardinals such as weakly compact cardinals, strongly unfoldable cardinals and Ramsey cardinals are characterized by the existence of “smaller” elementary embeddings for weak $\kappa$-models or $\kappa$-models of set theory. A weak $\kappa$-model of set theory is a set of size $\kappa$ containing $\kappa$ as an element and satisfying ${\rm ZFC}^-$, and a $\kappa$-model of set theory is a weak $\kappa$-model that is closed under ${<}\kappa$-sequences. As an example of such a characterization, $\kappa$ is weakly compact if and only if $2^{{<}\kappa}=\kappa$ and every $A\subseteq\kappa$ is contained in weak $\kappa$-model $M$ for which there is an elementary embedding $j:M\to N$ with critical point $\kappa$. The first author was motivated by obtaining indestructibility results for Ramsey cardinals to explore their little used elementary embeddings characterization, and the project ended up leading to general insights into the elementary embedding properties of smaller large cardinals [9], [10]. She introduced new large cardinal notions, the Ramsey-like cardinals, by generalizing the key properties of the elementary embeddings characterizing Ramsey cardinals. One type of Ramsey-like cardinals are the $\alpha$-iterable cardinals (for $1<\alpha<\omega_1$) which form a hierarchy between weakly compact and Ramsey cardinals. Another type are the strongly Ramsey cardinals which strengthen Ramsey cardinals but are weaker than measurable cardinals. Ramsey cardinals and $\alpha$-iterable cardinals are characterized by the existence of iterations of elementary embeddings on weak $\kappa$-models while strongly Ramsey cardinals are characterized by the existence of embeddings of $\kappa$-models, making them particularly suitable for standard indestructibility arguments which rely on the ${<}\kappa$-closure of the embedding source. We implement standard techniques to prove indestructibility results for strongly Ramsey cardinals and develop a uniform set of novel techniques to work with embeddings on weak $\kappa$-models and iterations of such embeddings.

Theorem:

Ramsey cardinals, $\alpha$-iterable cardinals, and strongly Ramsey
cardinals $\kappa$ are indestructible by:
- small forcing,
- the canonical forcing of the ${\rm GCH}$,
- the forcing to add a fast function on $\kappa$,
- the forcing to add a slim $\kappa$-Kurepa tree.
If $\kappa$ is a Ramsey, $\alpha$-iterable, or a strongly Ramsey cardinal, then there is a forcing extension in which $\kappa$ becomes indestructible by the forcing ${\rm Add}(\kappa,\theta)$ for every cardinal $\theta$.

Corollary:

The ${\rm GCH}$ can be forced to hold at a Ramsey, $\alpha$-iterable, or a strongly Ramsey cardinal.
The ${\rm GCH}$ can be forced to fail at a Ramsey, $\alpha$-iterable, or a strongly Ramsey cardinal.
If $\kappa$ is strongly Ramsey, then there is a forcing extension preserving this in which is not ineffable since it has slim-Kurepa tree.

If $\kappa$ is Ramsey, then there is a forcing extension in which the Ramseyness of $\kappa$ is destroyed but it remains virtually Ramsey.

[1] A. Lévy and R. M. Solovay, “Measurable cardinals and the continuum hypothesis,” Israel J. Math., vol. 5, pp. 234-248, 1967.
[Bibtex]

@article {levysolovay:ch,
AUTHOR = {L{\'e}vy, Azriel and Solovay, Robert M.},
TITLE = {Measurable cardinals and the continuum hypothesis},
JOURNAL = {Israel J. Math.},
FJOURNAL = {Israel Journal of Mathematics},
VOLUME = {5},
YEAR = {1967},
PAGES = {234--248},
ISSN = {0021-2172},
MRCLASS = {02.68},
MRNUMBER = {0224458 (37 \#57)},
MRREVIEWER = {G. Fodor},
}

[2] T. K. Menas, “Consistency results concerning supercompactness,” Trans. Amer. Math. Soc., vol. 223, pp. 61-91, 1976.
[Bibtex]

@article {menas:indes,
AUTHOR = {Menas, Telis K.},
TITLE = {Consistency results concerning supercompactness},
JOURNAL = {Trans. Amer. Math. Soc.},
FJOURNAL = {Transactions of the American Mathematical Society},
VOLUME = {223},
YEAR = {1976},
PAGES = {61--91},
ISSN = {0002-9947},
MRCLASS = {02K35 (02H13 02K05)},
MRNUMBER = {0540771 (58 \#27488)},
MRREVIEWER = {Jacques Stern},
}

[3] M. Gitik, “On measurable cardinals violating the continuum hypothesis,” Ann. Pure Appl. Logic, vol. 63, iss. 3, pp. 227-240, 1993.
[Bibtex]

@article {gitik:measurablenotCH,
AUTHOR = {Gitik, Moti},
TITLE = {On measurable cardinals violating the continuum hypothesis},
JOURNAL = {Ann. Pure Appl. Logic},
FJOURNAL = {Annals of Pure and Applied Logic},
VOLUME = {63},
YEAR = {1993},
NUMBER = {3},
PAGES = {227--240},
ISSN = {0168-0072},
CODEN = {APALD7},
MRCLASS = {03E55 (03E50)},
MRNUMBER = {1237232 (94m:03086)},
MRREVIEWER = {James Baumgartner},
DOI = {10.1016/0168-0072(93)90149-8},
URL = {http://dx.doi.org/10.1016/0168-0072(93)90149-8},
}

[4] R. Laver, “Making the supercompactness of $\kappa $ indestructible
under $\kappa $-directed closed forcing,” Israel J. Math., vol. 29, iss. 4, pp. 385-388, 1978.
[Bibtex]

@article {laver:supercompact,
AUTHOR = {Laver, Richard},
TITLE = {Making the supercompactness of {$\kappa $} indestructible
under {$\kappa $}-directed closed forcing},
JOURNAL = {Israel J. Math.},
FJOURNAL = {Israel Journal of Mathematics},
VOLUME = {29},
YEAR = {1978},
NUMBER = {4},
PAGES = {385--388},
ISSN = {0021-2172},
MRCLASS = {02K35 (02K05)},
MRNUMBER = {0472529 (57 \#12226)},
MRREVIEWER = {John Hickman},
}

[5] J. D. Hamkins and T. A. Johnstone, “Indestructible strong unfoldability,” Notre Dame J. Form. Log., vol. 51, iss. 3, pp. 291-321, 2010.
[Bibtex]

@article {hamkinsjohnstone:unfoldable,
AUTHOR = {Hamkins, Joel David and Johnstone, Thomas A.},
TITLE = {Indestructible strong unfoldability},
JOURNAL = {Notre Dame J. Form. Log.},
FJOURNAL = {Notre Dame Journal of Formal Logic},
VOLUME = {51},
YEAR = {2010},
NUMBER = {3},
PAGES = {291--321},
ISSN = {0029-4527},
MRCLASS = {03E55 (03E40)},
MRNUMBER = {2675684 (2011i:03050)},
MRREVIEWER = {Bernhard A. K{\"o}nig},
DOI = {10.1215/00294527-2010-018},
URL = {http://dx.doi.org/10.1215/00294527-2010-018},
}

[6] A. Kanamori, The higher infinite, Second ed., Berlin: Springer-Verlag, 2009.
[Bibtex]

@book {kanamori:higher,
AUTHOR = {Kanamori, Akihiro},
TITLE = {The higher infinite},
SERIES = {Springer Monographs in Mathematics},
EDITION = {Second},
NOTE = {Large cardinals in set theory from their beginnings,
Paperback reprint of the 2003 edition},
PUBLISHER = {Springer-Verlag},
ADDRESS = {Berlin},
YEAR = {2009},
PAGES = {xxii+536},
ISBN = {978-3-540-88866-6},
MRCLASS = {03E55 (01A55 01A60 03-02 03-03)},
MRNUMBER = {2731169},
}

[7] R. B. Jensen, “Measurable cardinals and the ${\rm GCH}$.” Providence, R.I.: Amer. Math. Soc., 1974, pp. 175-178.
[Bibtex]

@incollection {jensen:indestructibility,
AUTHOR = {Jensen, Ronald Bj{\"o}rn},
TITLE = {Measurable cardinals and the {${\rm GCH}$}},
BOOKTITLE = {Axiomatic set theory ({P}roc. {S}ympos. {P}ure {M}ath., {V}ol.
{XIII}, {P}art {II}, {U}niv. {C}alifornia, {L}os {A}ngeles,
{C}alif., 1967)},
PAGES = {175--178},
PUBLISHER = {Amer. Math. Soc.},
ADDRESS = {Providence, R.I.},
YEAR = {1974},
MRCLASS = {02K35 (02K05 02K25)},
MRNUMBER = {0369073 (51 \#5309)},
MRREVIEWER = {F. R. Drake},
}

[8] A. Beller, R. Jensen, and P. Welch, Coding the universe, Cambridge: Cambridge University Press, 1982, vol. 47.
[Bibtex]

@book {welch:codingpreserveramsey,
AUTHOR = {Beller, A. and Jensen, R. and Welch, P.},
TITLE = {Coding the universe},
SERIES = {London Mathematical Society Lecture Note Series},
VOLUME = {47},
PUBLISHER = {Cambridge University Press},
ADDRESS = {Cambridge},
YEAR = {1982},
PAGES = {i+353},
ISBN = {0-521-28040-0},
MRCLASS = {03-02 (03C62 03E45)},
MRNUMBER = {645538 (84b:03002)},
MRREVIEWER = {Thomas J. Jech},
}

[9] V. Gitman, “Ramsey-like cardinals,” The Journal of Symbolic Logic, vol. 76, iss. 2, pp. 519-540, 2011.
[Bibtex]

@ARTICLE {gitman:ramsey,
AUTHOR = {Victoria Gitman},
TITLE = {{R}amsey-like cardinals},
JOURNAL = {The Journal of Symbolic Logic},
VOLUME = {76},
YEAR = {2011},
NUMBER = {2},
PAGES = {519-540},
EPRINT={0801.4723},
PDF={http://boolesrings.org/victoriagitman/files/2011/08/ramseylikecardinals.pdf},
ISSN = {0022-4812},
CODEN = {JSYLA6},
MRCLASS = {03E55},
MRNUMBER = {2830415 (2012e:03110)},
MRREVIEWER = {Bernhard A. K{\"o}nig},
DOI = {10.2178/jsl/1305810762},
URL = {http://dx.doi.org/10.2178/jsl/1305810762},
}

[10] V. Gitman and P. D. Welch, “Ramsey-like cardinals II,” The Journal of Symbolic Logic, vol. 76, iss. 2, pp. 541-560, 2011.
[Bibtex]

@ARTICLE{gitman:welch,
AUTHOR= "Victoria Gitman and Philip D. Welch",
TITLE= "Ramsey-like cardinals {II}",
JOURNAL = {The Journal of Symbolic Logic},
VOLUME = {76},
YEAR = {2011},
NUMBER = {2},
PAGES = {541-560},
PDF={http://boolesrings.org/victoriagitman/files/2011/08/ramseylikecardinalsii.pdf},
EPRINT ={1104.4448},
ISSN = {0022-4812},
CODEN = {JSYLA6},
MRCLASS = {03E55},
MRNUMBER = {2830435 (2012e:03111)},
MRREVIEWER = {Bernhard A. K{\"o}nig},
DOI = {10.2178/jsl/1305810763},
URL = {http://dx.doi.org/10.2178/jsl/1305810763},
}

epub, mathjax and the iPad — another attempt

Peter Krautzberger — Mon, 17 Oct 2011 18:44:29 +0000

It’s a funny thing. I don’t even own an iPad. But a lot of people are interested in getting an epub file with mathjax working on the iPad.

Why is that? Well, as far as I could find out the …

Grigorieff forcing collapsing the continuum

Peter Krautzberger — Sun, 16 Oct 2011 05:19:20 +0000

This is a short technical post, more a note-to-self so that I know where to look this up if I ever need it again. It is also somewhat of a correction of something I said during my talk in Toronto …

Counting Birds…

François G. Dorais — Sat, 15 Oct 2011 17:35:30 +0000

I’ve been thinking a bit more about the issues brought up in Paris–Harrington and Typing and Santa Exists! These issues seem related to Borges’s ornithological argument for the existence of God:1 Cierro los ojos y veo una bandada de pájaros. La visión dura un segundo o acaso menos; no sé cuántos pájaros vi. ¿Era definido [...]

Santa Exists!

François G. Dorais — Sat, 15 Oct 2011 15:38:37 +0000

Here is a theorem that I like to mention during the first week when I teach set theory. Theorem. Santa exists! Proof. Consider the set $$S = \set{X : X \in X \lthen \text{Santa exists}}.$$ Then $$S \in S \liff (S \in S \lthen \text{Santa exists}).\tag{$*$}$$ The forward implication of $(*)$ gives that $$S \in [...]

A posting on wordpress-for-scientists

Peter Krautzberger — Sat, 15 Oct 2011 04:31:11 +0000

I just finished a long posting at the mailing list/google group WordPress For Scientist. This was spawned by today’s meeting with Sam and this week’s trouble with the papercite plugin. We really need to find a different solution.
…

Getting started…

François G. Dorais — Fri, 14 Oct 2011 19:32:59 +0000

As you probably noticed, I’ve started rolling out some research contents. I think I’ve got the basic structure that I want for the site. So I’m starting to think about minor tweaks and making things look pretty. Let me know you have some suggestions and tips!

Complete subspaces of a metric space

François G. Dorais — Fri, 14 Oct 2011 19:10:57 +0000

$\DeclareMathOperator{\add}{add}\DeclareMathOperator{\cof}{cof}\DeclareMathOperator{\cov}{cov}\DeclareMathOperator{\non}{non}\newcommand{\T}{\mathcal{T}}\newcommand{\M}{\mathcal{M}}$ Let $X$ be an arbitrary metric space. What are the possible structural properties for the ideal $\T(X)$ generated by the complete subspaces of $X$? How does the topology of $X$ affect $\cof(\T(X))$? $\cov(\T(X))$? What about the $\sigma$-ideal $\T_\sigma(X)$ generated by the complete subspaces of $X$? These questions are superficially very similar to questions asked [...]

Stationary strategies in Choquet games

François G. Dorais — Fri, 14 Oct 2011 04:46:41 +0000

The (strong) Choquet game on a topological space $X$ is played as follows. There are two players, Empty and Nonempty, who alternate turns for infinitely many rounds. On round $i$, Empty moves first, choosing a point $x_i$ and an open neighborhood $U_i$ of $x_i$ and, if $i \geq 1$, such that $U_i \subseteq V_{i-1}$ (the [...]

Paris-Harrington and Typing

François G. Dorais — Fri, 14 Oct 2011 04:44:34 +0000

I just read an interesting paper by Thomas Forster: The Paris–Harrington Theorem in an NF context [1]. This paper is a progress report on the status of the Paris–Harrington Theorem in Quine’s New Foundations (NF). Forster shows that the Paris–Harrington Theorem is at least consistent with NF, but leaves open the question whether it is [...]

A game-theoretic proof of Fraïssé’s Theorem

François G. Dorais — Fri, 14 Oct 2011 04:44:33 +0000

One of the important outcomes of the Fraïssé approach to model theory is that it is possible, in principle, to develop model theory from a strictly structural perspective, without ever mentioning formulas and their semantics. In practice, nobody really goes that far, but it is very useful to know about this radically different way to [...]

Oracle forcing Part I

KT — Thu, 13 Oct 2011 18:41:53 +0000

Introduction and definitions

Recently, Martin Goldstern and I have been studying oracle-cc and oracle-proper forcing, which was introduced by Shelah in his paper numbered 100 (proper forcing was also introduced in this paper) which is creatively and descriptively titled Independence Results.

Before I get to what these things are, a brief note on why. Oracle forcings in all variants have the property that the ground model reals remain of second category in the extension. They also have the Omitting Types Theorem or “what is forced to be dead, stays dead” (details can be found in Proper and Improper Forcing, Chapter IV).

There are some applications of oracle-cc forcing (e.g. Geschke-Kojman or Mildenberger) but so far the only application of oracle-proper forcing is the one given in the original Shelah 100 paper. Shelah uses an iteration of oracle-proper forcings to come up with a model in which there is a universal linear order at $\aleph_1$ and CH fails. This application is important, the method of getting the universal is deceptively simple and so the technique is worth studying.

Martin and I started by studying Uri Abrahams notes from a tutorial given at the Logic Colloquium in Paris in 2010. The definitions and the first conceptual ideas that I will state stem from those notes, but are ultimately the same as the original Shelah concepts. Then we went on to study how these notions of oracle forcing fit into the framework of familiar forcing categories such as proper, ccc, $\sigma$-closed, $\sigma$-centred. I will try to relate this work here.

Definition: A sequence $\bar{M} = \langle M_\delta : \delta < \omega_1,$ limit $\rangle$ is an oracle (strong oracle) iff

for each $\delta$ $M_\delta$ is a countable elementary submodel of some $H(\xi)$ for sufficiently large $\xi$ such that $\delta \in M_\delta$ and $M_\delta \vDash \delta$ is countable
for any $A \subseteq \omega_1$ the set of all $\delta \in \omega_1$ such that $A \cap \delta \in M_{\delta}$ is stationary (club).

To see that the existence of oracles is equivalent to $\diamondsuit_{\aleph_1}$, first let $$\bar{A} = \langle A_{\alpha} : \alpha \in \omega_1\rangle$$
be a $\diamondsuit_{\aleph_1}$-sequence and for each limit $\delta \in \omega_1$ let $M_{\delta}$ be the collapse of a countable elementary subset of $H(\xi)$ containing $A_\delta$ (for some $\xi$ sufficiently large) such that $\delta \in M_{\delta}$.

The opposite direction is shown by taking as $A_\delta$ all subsets of $\delta$ in $M_\delta$ (and $A_\delta = \emptyset$ for $\delta$ a successor ordinal). This is a easily seen to be a $\diamondsuit^-$ sequence which is equivalent to $\diamondsuit$ by a theorem of Kunen.

Definition: An oracle (strong oracle) $\bar{M}^\prime = \langle M^\prime_{\delta} : \delta \in \omega_1, \text{ limit}\rangle$ is a proper extension of an oracle (strong oracle) $\bar{M}$, denoted $\bar{M} \trianglelefteq \bar{M^\prime}$, if for a club subset of $\delta \in \omega_1$

$M_{\delta} \in M^\prime_{\delta}$ and $M_{\delta} \subseteq M^\prime_{\delta}$
$M^\prime_{\delta} \vDash |M_{\delta}| = \aleph_0$.

Shelah proves that given any oracle, we may find another oracle which is a proper extension of it. In fact, this process may be repeated $\aleph_1$-many times.

For an oracle $\bar{M}$, and $A \subseteq \omega_1$ let $I_{\bar{M}}(A)$ be the set of non-zero limit ordinals $\delta \in \omega_1$ such that $A \cap \delta \in M_\delta$. Let $D_{\bar{M}}$ be the filter which is generated by $I_{\bar{M}}(A)$ for all $A \subseteq \omega_1$. The filter $D_{\bar{M}}$ is normal and contains the club filter; proofs of these facts can be found in Abrahams notes.

Definition: Let $\bar{M}$ be an oracle and let $P$ be a forcing poset such that the universe of $P$ is $\omega_1$. Let $D(P)$ be the set of all $\delta \in \omega_1$ such that for all $E \in M_\delta$, $E$ is pre-dense in $P \upharpoonright \delta$ implies $E$ is pre-dense in $P$. Then $P$ is $\bar{M}$-cc iff $D(P) \in D_{\bar{M}}$.

A simpler way of writing this is that $D(P) = \{\delta \in \omega_1 : P_\delta \preceq_{M_\delta} P\} \in D_{\bar{M}}$.

The fact that any $\bar{M}$-cc satisfies the ccc can be found in Abrahams notes.

Definition: Let $\bar{M}$ be an oracle and let $P$ be a forcing poset such that the universe of $P$ is $\omega_1$. For any $p \in P$ let $D_p(P)$ be the set of all $\delta \in \omega_1$ such that there exists $q_\delta \leq p$ for all $E \in M_\delta$, $E$ is pre-dense in $P \upharpoonright \delta$ implies $E$ is pre-dense in $P$ below $q_\delta$. Then $P$ is $\bar{M}$-proper iff $D_p(P) \in D_{\bar{M}}$.

Such $q_\delta$ are called $(\delta, M_\delta)$-generic.

As Abraham points out, we can combine the sets $D_p(P)$ for $p \in P$ to obtain a single set $D(P)$ such that if $\delta \in D(P)$ for each $p \in P \upharpoonright \delta$ there is $q \leq p$ which is $(\delta, M_\delta)$-generic.

In order to work with $P$ which do not have universe $\omega_1$ Abraham provides an equivalent definition. Let $K \prec H(\lambda)$ have size $\aleph_1$ such that $P \in K$, and let $s_\alpha : K_\alpha \rightarrow \bar{K_\alpha}$ be the collapse function. Denote $\bar{P}_\alpha = s_\alpha(P)$, the copy of $P$ in $\bar{K_\alpha}$. Let $\delta \in \omega_1$ be such that $\bar{K_\alpha} \in M_\delta$. We say that $q \in P$ is $(K_\alpha, M_\delta)$-generic if for every $X \in M_\delta$ dense in $\bar{P}_\alpha$ we have $s^{-1}(X) \subseteq P \cap K_\alpha$ is predense in $P$ below $q$.

Theorem (Abraham):

$P$ is $\bar{M}$-proper iff for every $K$ and $\langle K_\alpha : \alpha \in \omega_1 \rangle$ as above, there is $I(P) \in D_{\bar{M}}$ such that for each $\delta \in I(P)$ with $\bar{K_\delta} \in M_\delta$ and for every $p \in P \cap K_\delta$ there is $q \leq p$ which is $(K_\delta, M_\delta)$-generic.

Notation: We say that a forcing is uberoracle-cc (uberoracle-proper) if and only if it is $\bar{M}$-cc ($\bar{M}$-proper) for every oracle $\bar{M}$. Also, we abbreviate the property “$\bar{M}$-cc ($\bar{M}$-proper) for a strong oracle $\bar{M}$” by strong $\bar{M}$-cc (strong $\bar{M}$-proper).

The reason that we define uberoracle-proper is that the only known application of oracle-proper forcing has this stronger property.

5th Young Set Theory Workshop

KT — Wed, 12 Oct 2011 13:58:37 +0000

Hurrah, the next YST workshop has been announced!

The official website

It will take place at the beautiful location of the CIRM at the Luminy campus near Marseilles. The dates: 30 April – 4 May 2012.

‘Young’ is relative and not well-defined, so hope to see lots of colleagues there!

What is the theory ZFC without power set?

Victoria Gitman — Sun, 09 Oct 2011 21:13:01 +0000

V. Gitman, J. D. Hamkins, and T. A. Johnstone, “What is the theory ZFC without power set?.” (Submitted)

PDF Citation arχiv

@ARTICLE{zfcminus:gitmanhamkinsjohnstone,
AUTHOR= {Victoria Gitman and Joel David Hamkins and Thomas A. Johnstone},
TITLE= {What is the theory {ZFC} without power set?},
NOTE= {Submitted},
PDF={http://boolesrings.org/victoriagitman/files/2011/10/ZFC-.pdf},
EPRINT={1110.2430}}

This is joint work with Joel David Hamkins and Thomas Johnstone.

Set theory without the power set axiom is used in arguments and constructions throughout the subject and is usually described simply as having all the axioms of $\rm{ZFC}$ except for the power set axiom. This theory arises frequently in the large cardinal theory of iterated ultrapowers, for example, and perhaps part of its attraction is an abundance of convenient natural models, including $\langle H_\kappa,{\in}\rangle$ for any uncountable regular cardinal $\kappa$, where $H_\kappa$ consists of sets with hereditary size less than $\kappa$. When prompted, many set theorists offer a precise list of axioms: extensionality, foundation, pairing, union, infinity, separation, replacement and choice. Let us denote by $\rm{ZFC}{-}$ the theory having the axioms listed above with the axiom of choice taken to mean Zermelo’s well-ordering principle, which then implies Zorn’s Lemma as well as the existence of choice-functions. These alternative formulations of choice are not all equivalent without the power set axiom as is proved by Zarach in [1], in particular there are models of $\rm{ZF}^-$ in which choice-functions exist but Zermelo’s well-ordering principle fails. Zarach initiated the program of establishing unintuitive consequences of set theory without power set, which we carry on in this article.

In this article, we shall prove that this formulation of set theory without the power set axiom is weaker than may be supposed and is inadequate to prove a number of basic facts that are often desired and applied in its context. Specifically, we shall prove that the following behavior can occur with $\rm{ZFC}{-}$ models.

(Zarach, [2]) There are models of $\rm{ZFC}{-}$ in which the countable union of countable sets is not necessarily countable, indeed, in which $\omega_1$ is singular, and hence the collection axiom scheme fails.
(Zarach [2]) There are models of $\rm{ZFC}{-}$ in which every set of reals is countable, yet $\omega_1$ exists.
There are models of $\rm{ZFC}{-}$ in which for every $n<\omega$, there is a set of reals of size $\aleph_n$, but there is no set of reals of size $\aleph_\omega$.
The Łós ultrapower theorem can fail for $\rm{ZFC}{-}$ models.
- There are models $M\models\rm{ZFC}{-}$ with an $M$-normal measure $\mu$ on a cardinal $\kappa$ in $M$, for which the ultrapower by $\mu$, using functions in $M$, is well-founded, but the ultrapower map is not elementary.
- Such violations of Łós can arise even with internal ultrapowers on a measurable cardinal $\kappa$, where $P(\kappa)$ exists in $M$ and $\mu\in M$.
- There is $M\models\rm{ZFC}{-}$ in which $P(\omega)$ exists in $M$ and there are ultrafilters $\mu$ on $\omega$ in $M$,
  but no such $M$-ultrapower map is elementary.
The Gaifman theorem [3] can fail for $\rm{ZFC}{-}$ models.
- There are $\Sigma_1$-elementary cofinal maps $j:M\to N$ of transitive $\rm{ZFC}{-}$ models, which are not elementary.
- There are elementary maps $j:M\to N$ of transitive $\rm{ZFC}{-}$ models, such that the canonical cofinal restriction $j:M\to \bigcup j''M$ is not elementary.
Seed theory arguments can fail for $\rm{ZFC}{-}$ models. There are elementary embeddings $j:M\to N$ of transitive $\rm{ZFC}{-}$ models and sets $S\subseteq\bigcup j'' M$ such that the seed hull $\mathbb X_S=\{j(f)(s)\mid s\in [S]^{{\lt}\omega}, f\in M\}$} of $S$ is not an elementary submodel of $N$.
The collection of formulas that are provably equivalent in $\rm{ZFC}{-}$ to a $\Sigma_1$-formula or a $\Pi_1$-formula is not closed under bounded quantification.

Nevertheless, these deficits of $\rm{ZFC}{-}$ are completely repaired by strengthening it to the theory $\rm{ZFC}^-$, obtained by using collection rather than replacement in the axiomatization above.

[1] A. Zarach, “Unions of ${\rm ZF}^{-}$-models which are themselves
${\rm ZF}^{-}$-models.” Amsterdam: North-Holland, 1982, vol. 108, pp. 315-342.
[Bibtex]

@incollection {zarach:unions_of_zfminus_models,
AUTHOR = {Zarach, Andrzej},
TITLE = {Unions of {${\rm ZF}^{-}$}-models which are themselves
{${\rm ZF}^{-}$}-models},
BOOKTITLE = {Logic {C}olloquium '80 ({P}rague, 1980)},
SERIES = {Stud. Logic Foundations Math.},
VOLUME = {108},
PAGES = {315--342},
PUBLISHER = {North-Holland},
ADDRESS = {Amsterdam},
YEAR = {1982},
MRCLASS = {03C62 (03E35)},
MRNUMBER = {673801 (84h:03086)},
MRREVIEWER = {M. Dubiel},
}

[2] A. M. Zarach, “Replacement $\nrightarrow$ collection.” Berlin: Springer, 1996, vol. 6, pp. 307-322.
[Bibtex]

@incollection {Zarach1996:ReplacmentDoesNotImplyCollection,
AUTHOR = {Zarach, Andrzej M.},
TITLE = {Replacement {$\nrightarrow$} collection},
BOOKTITLE = {G\"odel '96 ({B}rno, 1996)},
SERIES = {Lecture Notes Logic},
VOLUME = {6},
PAGES = {307--322},
PUBLISHER = {Springer},
ADDRESS = {Berlin},
YEAR = {1996},
MRCLASS = {03E30 (03E35)},
MRNUMBER = {1441120 (98g:03120)},
}

[3] H. Gaifman, “Elementary embeddings of models of set-theory and certain
subtheories.” Providence R.I.: Amer. Math. Soc., 1974, pp. 33-101.
[Bibtex]

@incollection {gaifman:ultrapowers,
AUTHOR = {Gaifman, Haim},
TITLE = {Elementary embeddings of models of set-theory and certain
subtheories},
BOOKTITLE = {Axiomatic set theory ({P}roc. {S}ympos. {P}ure {M}ath., {V}ol.
{XIII}, {P}art {II}, {U}niv. {C}alifornia, {L}os {A}ngeles,
{C}alif., 1967)},
PAGES = {33--101},
PUBLISHER = {Amer. Math. Soc.},
ADDRESS = {Providence R.I.},
YEAR = {1974},
MRCLASS = {02K15 (02H13)},
MRNUMBER = {0376347 (51 \#12523)},
MRREVIEWER = {L. Bukovsky},
}

Hello Booles’ Rings!

François G. Dorais — Sun, 09 Oct 2011 19:49:21 +0000

I’m glad to be a member of Booles’ Rings! Please be patient during the construction process…

Formal proofs are our democracy

Peter Krautzberger — Sat, 08 Oct 2011 16:11:22 +0000

Reading papers can lead to horrible acts. Today, I felt like mutilating a famous quote.

Many forms of communicating mathematics have been tried and will be tried in this world of sin and woe. No one pretends that formal
…

Flat ultrafilters follow-up

Peter Krautzberger — Thu, 29 Sep 2011 16:28:18 +0000

There are a couple of reasons to write a quick follow-up post to the notes for my talk on flat ultrafilters.

Small addenda included

I updated the notes a little. Nothing much, some bad pictures and some comments.

Video …

Flat Ultrafilters (Michigan Logic Seminar Sept 21, 2011)

Peter Krautzberger — Sun, 25 Sep 2011 19:36:37 +0000

Remember how our about page says that Booles’ Rings is about best practices for an acacdemic homepage? Ok, let’s try one: making notes to talks available.

Some introductory remarks

Skip this section if you only want mathematics.

Wednesday, I gave …

Hindman’s Theorem, partial semigroups and some of my most lacking intuitions (part 7)

Peter Krautzberger — Thu, 22 Sep 2011 01:27:16 +0000

Hm, my writing process is slowing down a little (and on top of that I forgot to publish this draft) and there are other posts that I really want to write. I’m not really sure how I will proceed, but
…

Hindman’s Theorem, partial semigroups and some of my most lacking intuitions (part 6)

Peter Krautzberger — Thu, 15 Sep 2011 16:01:58 +0000

I know, I know, it’s part 6 already. Last time I finally formulated the Central Sets Theorem. This part will just be a small bridge. But at least you’ll finally know why on earth I am writing this, i.e., what
…

Hindman’s Theorem, partial semigroups and some of my most lacking intuitions (part 5)

Peter Krautzberger — Thu, 08 Sep 2011 23:01:29 +0000

Last time, I left you hanging — I promised the Central Sets Theorem, but only bothered you with some more stuff on partial semigroup, i.e., condensations. Let me make it up to you.

The Central Sets Theorems

So what …

Hindman’s Theorem, partial semigroups and some of my most lacking intuitions (part 4)

Peter Krautzberger — Wed, 07 Sep 2011 14:56:06 +0000

Well, I hope you didn’t miss me while I was on my first summer vacation in three years. So let’s continue this series. If you remember, part 3 consisted mainly of the observation that FS-sets have a partial semigroup structure
…

Hindman’s Theorem, partial semigroups and some of my most lacking intuitions (part 3)

Peter Krautzberger — Thu, 25 Aug 2011 21:00:50 +0000

Ok, time for part 3. We’re not close to an end but I must apologize that I won’t be able to post in the next week. But let’s recap. In the first part I simply explained why semigroups are not
…

Hindman’s Theorem, partial semigroups and some of my most lacking intuitions (part 2)

Peter Krautzberger — Wed, 24 Aug 2011 16:21:27 +0000

Yesterday I finally started this short series with some small thoughts regarding partial semigroups. If you don’t remember, all I did yesterday was to explain why semigroups are not, in general, partition regular using a simple partition of $\mathbb{N}$. I
…

Hindman’s Theorem, partial semigroups and some of my most lacking intuitions (part 1)

Peter Krautzberger — Tue, 23 Aug 2011 13:21:59 +0000

If you remember, I mentioned that I was working on a post on some research and it was getting out of hand. Well, it is still not finished, but long enough to start posting a series of posts.

It’s no …

What’s in a name

Peter Krautzberger — Mon, 22 Aug 2011 18:46:27 +0000

This is a slightly idolized recollection how we came to adopt the name Booles’ Rings.

When Sam and I began looking seriously into the idea of wordpress for mathematicians we also started thinking about names. I immediately thought of one …

A short reflection on google+

Peter Krautzberger — Sat, 20 Aug 2011 01:40:48 +0000

Today I spent a lot of time on writing a longer piece about some mathematics. It grew completely out of proportion and it’ll take me a few more ~~minutes~~ hours to finish.

Other things have been keeping me busy but …

WordPress in a VM

Peter Krautzberger — Wed, 10 Aug 2011 21:08:14 +0000

When you’re using wordpress, it comes in handy to have an installation to play around with. When I first joined the WordPress for Scientists Google Group, I promised to write a tutorial. Well, it took me a bit to …

About page for Booles’ Rings

Peter Krautzberger — Mon, 08 Aug 2011 01:30:32 +0000

I just published a first draft of an about page for Booles’ Rings.

If you have any comments, please leave them here.…

Why markdown, not $\LaTeX$?

Peter Krautzberger — Thu, 04 Aug 2011 01:03:32 +0000

Coming from $\LaTeX$ and its text-editor driven writing style, working with wordpress can seem a step back — after all, most people, if they get excited about $\TeX$, very much despise the wysiwyg approach to writing.

Its greatest strength, its …

Hello subjects and fans!

KT — Sun, 31 Jul 2011 16:08:56 +0000

Welcome to Booles’ Rings. Woo wordpress!

And now the continuation…

Peter Krautzberger — Sat, 30 Jul 2011 16:27:37 +0000

Alright, Booles’ Rings is set up and I’m trying to import my old posts from http://peter.krautzberger.info. Jekyll was fun, but WordPress has more potential. So let’s start something new.

…

Shelah’s Model without P-points– part 9

admin — Tue, 19 Jul 2011 04:00:00 +0000

Read more about this series at the first post.

Part 9: the main lemma ctd.

In short:

Destroying P-points in any further $\omega^\omega$-bounding extensions.
- Second and final part of the proof.

Part 9 as PDF

Part 9 …

The Mathematician’s Homepage — can it be more?

admin — Sun, 17 Jul 2011 04:00:00 +0000

Yesterday I went on a rant about Keith Devlin’s homepage and twitter. That was, like all rants, a little unfair on him. I guess idols are always more disappointing. However, although the rant was triggered by that one tweet it
…

Dear Keith Devlin

admin — Sat, 16 Jul 2011 04:00:00 +0000

Dear Keith Devlin,

I love your work. Really, I do. So please don’t take this rant the wrong way. I promise I’ll shut up after this one rant, I just really need to get this off my chest.

#begin rant…

Why I back Relatively Prime

admin — Thu, 14 Jul 2011 04:00:00 +0000

On of the pleasures of running mathblogging.org is that you get to find out about immensely creative people working on new ways to make use of the web for mathematics.

ACME Science

Only a few weeks ago, I rediscovered Samuel …

Epub and mathematics

admin — Wed, 13 Jul 2011 04:00:00 +0000

A while ago Martin Fenner had written about the BeyondPDF workshop and released his epub export plugin for wordpress and started the WordPress For Scientist Google group.

I had joined that group right away and had wanted to do …

Shelah’s Model without P-points– part 8

admin — Mon, 11 Jul 2011 04:00:00 +0000

Read more about this series at the first post.

Part 8: the main lemma

In short:

Destroying P-points in any further $\omega^\omega$-bounding extensions.
- First part of the proof.

Part 8 as PDF

Part 8 as Xournal-source…

Finding regular insertion encodings for permutation classes

Vince Vatter — Sun, 10 Jul 2011 16:00:14 +0000

We describe a practical algorithm which computes the accepting automaton for the insertion encoding of a permutation class, whenever this insertion encoding is regular. This algorithm is implemented in the accompanying Maple package INSENC, which can automatically compute the rational generating functions for such classes.

preprint pdf

Small permutation classes

Vince Vatter — Fri, 08 Jul 2011 16:00:06 +0000

Proceedings of the London Mathematical Society, to appear.

We establish a phase transition for permutation classes (downsets of permutations under the permutation containment order): there is an algebraic number $\kappa$, approximately $2.20557$, for which there are only countably many permutation classes of growth rate (Stanley-Wilf limit) less than $\kappa$ but uncountably many permutation classes of growth rate $\kappa$, answering a question of Klazar. We go on to completely characterize the possible sub-$\kappa$ growth rates of permutation classes, answering a question of Kaiser and Klazar. Central to our proofs are the concepts of generalized grid classes (introduced herein), partial well-order, and atomicity (also known as the joint embedding property).

download from the journal (free-access link)
preprint pdf
source: zip

Shelah’s Model without P-points– part 7

admin — Fri, 01 Jul 2011 04:00:00 +0000

My apologies for the blogging hiatus. Let’s continue.

Read more about this series at the first post.

Part 7: switching filters

In short:

From $u$ to $\{ \omega \} \otimes u$.

Part 7 as PDF

Part 7 …

Shelah’s Model without P-points– part 5

admin — Thu, 02 Jun 2011 04:00:00 +0000

Read more about this series at the first post.

Part 5: properness

In short:

Basic Properties:
- Grigorieff forcing and the Sacks variant are proper (part 1)

Part 5 as PDF

Part 6 as Xournal-source…

Shelah’s Model without P-points– part 6

admin — Thu, 02 Jun 2011 04:00:00 +0000

Read more about this series at the first post.

Part 6: properness ctd.

In short:

Basic Properties:
- Grigorieff forcing and the Sacks variant are proper (part 2)

Part 6 as PDF

Part 6 as Xournal-source…

Counting $({\bf 3}+{\bf 1})$-avoiding permutations

Vince Vatter — Wed, 01 Jun 2011 16:00:25 +0000

European Journal of Combinatorics, to appear.
With Mike Atkinson and Bruce Sagan.

A poset is $({\bf 3}+{\bf 1})$-free if it contains no induced subposet isomorphic to the disjoint union of a 3-element chain and a 1-element chain. These posets are of interest because of their connection with interval orders and their appearance in the $({\bf 3}+{\bf 1})$-free Conjecture of Stanley and Stembridge. The dimension 2 posets $P$ are exactly the ones which have an associated permutation $\pi$ where $i\prec j$ in $P$ if and only if $i<j$ as integers and $i$ comes before $j$ in the one-line notation of $\pi$. So we say that a permutation $\pi$ is $({\bf 3}+{\bf 1})$-free or $({\bf 3}+{\bf 1})$-avoiding if its poset is $({\bf 3}+{\bf 1})$-free. This is equivalent to $\pi$ avoiding the permutations $2341$ and $4123$ in the language of pattern avoidance. We give a complete structural characterization of such permutations. This permits us to find their generating function.

The sequence found in the paper is A165531 in the OEIS.

preprint pdf

On points drawn from a circle

Vince Vatter — Wed, 01 Jun 2011 16:00:19 +0000

With Steve Waton

Choose n points from the unit circle, no two with the same $x$-coordinate or $y$-coordinate. Label these points $1$ to $n$ by height, reading bottom to top, and record these labels reading left to right. This operation produces a permutation. For example, the set of points shown below on the left gives the permutation $45312$, the plot of which is shown on the right.
We characterise and enumerate the permutations that arise from this correspondence.

preprint pdf

Shelah’s Model without P-points– part 4

admin — Mon, 30 May 2011 04:00:00 +0000

Read more about this series at the first post.

Part 4: $\omega^\omega$-bounding

In short:

Basic Properties:
- Grigorieff forcing and the Sacks variant are $\omega^\omega$-bounding

Part 4 as PDF

Part 4 as Xournal-source…

Shelah’s Model without P-points– part 3

admin — Sat, 28 May 2011 04:00:00 +0000

Read more about this series at the first post.

Part 3: More Strategy and Shelah’s “crucial fact”

In short:

Basic Properties:
a strategy for $\omega^\omega$-bounding — continued.
The “crucial fact” for Grigorieff and Sacks forcing (that’s what Shelah calls

Maximal independent sets and separating covers

Vince Vatter — Sun, 01 May 2011 16:00:50 +0000

American Mathematical Monthly 118 (2011), 418–423.

The answer to the fourth problem in MGIII (what is the largest integer which is the product of positive integers with sum $n$?) turns out to be an inverse to the solution of the last problem (Katona’s problem of determining the maximum number of subsets in a separating family with $n$ elements). We give a combinatorial explanation for this relationship, via Moon and Moser’s answer to another question (how many maximal independent sets can a graph on $n$ vertices have?).

download from the journal (subscription required)
preprint pdf

Subclasses of the separable permutations

Vince Vatter — Fri, 29 Apr 2011 16:00:03 +0000

Bulletin of the London Mathematical Society, to appear.
With Michael Albert and Mike Atkinson.

We prove that all subclasses of the separable permutations not containing $Av(231)$ or a symmetry of this class have rational generating functions. Our principal tools are partial well-order, atomicity, and the theory of strongly rational permutation classes introduced here for the first time.

download from the journal (free-access link)
preprint pdf
source: source

On convex permutations

Vince Vatter — Sat, 12 Feb 2011 00:09:26 +0000

Discrete Mathematics 311 (2011), 715–722.
With Michael Albert, Steve Linton, Nik Ruškuc, and Steve Waton.

A selection of points drawn from a convex polygon, no two with the same vertical or horizontal coordinate, yields a permutation in a canonical fashion. We characterise and enumerate those permutations which arise in this manner and exhibit some interesting structural properties of the closed class they form. We conclude with a permutation analogue of the celebrated Happy Ending Problem.

download from the journal (subscription required)
preprint pdf

Simple extensions of combinatorial structures

Vince Vatter — Tue, 21 Dec 2010 17:00:08 +0000

Mathematika 57 (2011), 193–214.
With Robert Brignall and Nik Ruškuc.

An interval in a combinatorial structure $S$ is a set $I$ of points which relate to every point from $S\setminus I$ in the same way. A structure is simple if it has no proper intervals. Every combinatorial structure can be expressed as an inflation of a simple structure by structures of smaller sizes — this is called the substitution (or modular) decomposition. In this paper we prove several results of the following type: An arbitrary structure $S$ of size $n$ belonging to a class $\mathcal{C}$ can be embedded into a simple structure from $\mathcal{C}$ by adding at most $f(n)$ elements.

We prove such results when $\mathcal{C}$ is the class of all tournaments, graphs, permutations, posets, digraphs, oriented graphs and general relational structures containing a relation of arity greater than 2. The function $f(n)$ in these cases is $2$, $\lceil \log_2(n+1)\rceil$, $\lceil (n+1)/2\rceil$, $\lceil (n+1)/2\rceil$, $\lceil \log_4(n+1)\rceil$, $\lceil \log_3(n+1)\rceil$ and $1$, respectively. In each case these bounds are best possible.

download from the journal (free-access link)
preprint pdf
source

On partial well-order for monotone grid classes of permutations

Vince Vatter — Fri, 16 Jul 2010 16:00:17 +0000

Order 28 (2011), 193–199.
With Steve Waton.

A monotone grid class is a permutation class (i.e., a downset of permutations under the containment order) defined by local monotonicity conditions. We give a simplified proof of a result of Murphy and Vatter that monotone grid classes of forests are partially well-ordered.

download from the journal (subscription required)
preprint pdf
source

Working with ranges in Maple

Vince Vatter — Sun, 16 May 2010 00:59:46 +0000

As with most things in Maple, ranges are under-developed and poorly documented, but nonetheless quite useful.Technically a range is an expression of the form

expression..expression

Ranges are great for summing series

> sum(x^i, i=1..infinity);
                                        x
                                    - -----
                                      x - 1

or specifying sublists

> a := [1,2,3,4,5,6,7,8,9];
                       a := [1, 2, 3, 4, 5, 6, 7, 8, 9]

> op(3..7, a);
                                 3, 4, 5, 6, 7

or building sets:

> {$3..12};
                       {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(here the $ tells Maple to expand the range into an expression sequence — note that you cannot expand an infinite range). For some odd reason, ranges can’t be used in for loops:

> for i in 2..9 do
>   print(i);
> od;
                                       2

                                       9

Below I have included a few things that I wish Maple explained in their documentation of ranges.

Getting the endpoints

A range is a datatype with two ‘operands’, but for some reason you can’t access these in the normal way:

> r := 2..infinity;
                              r := 2 .. infinity

> r[1];
                              (2 .. infinity)[1]

Instead, you have to use the op procedure:

> op(1, r);
                                       2

> op(2, r);
                                   infinity

The membership problem

Often I wonder why Maple hates me so much. Why can’t I just ask if a number is in a range?

> member(3, 1..10);
Error, invalid input: member received 1 .. 10, which is not valid for its 2nd
argument, s

Sure, in this case I could convert the range to a set:

> member(3, {$1..10});
                                     true

But that’s not going to help much with an infinite range, because Maple will not expand infinite ranges into expression sequences (what the $ is doing above). Even worse than simply not working, using this approach with infinite ranges gives you garbage instead of an error message:

> member(3, {$1..infinity});
                                     false

So if you need to test range membership, you have to write your own procedure:

member_range := proc(x::integer, r::range)
  if (x >= op(1, r)) and (x  member_range(3,1..infinity);
                                     true

> member_range(-3,1..infinity);
                                     false

Sure, that’s not the hardest code to write, but why can’t Maple just overload the member function?

Intersecting ranges

Another glaring omission is range intersection:

> (-4..infinity) intersect (-infinity..10)
>

Since it refuses to do anything at all, we’ll need to replace the intersect command. This isn’t hard, but still…

intersect_range := proc(r1::range, r2::range)
  return(max(op(1, r1), op(1, r2))..min(op(2, r1), op(2, r2)));
end:

> intersect_range(-4..infinity, -infinity..10);
                                   -4 .. 10

> intersect_range(-4..infinity, 10..infinity);
                                10 .. infinity

A bit surprisingly, this even works when we don’t explicitly mention infinity:

> intersect_range(-4.., 10..);
                                10 .. infinity

The only drawback with this approach is that empty ranges don’t get a canonical form:

> intersect_range(-4..3, 8..10);
                                    8 .. 3

Fortunately, if you apply a $ to the range 8..3, Maple is smart enough to return nothing at all.

Because ranges are restricted to a single ‘connected component’, there is no way to implement the union or minus commands.

Suffixes of lists

At some point between Maple 11 and Maple 13, a handy trick was introduced. To take a suffix of a list, you no longer need to specify the end of the range, for example,

> a := [1,2,3,4,5,6,7,8,9];
                       a := [1, 2, 3, 4, 5, 6, 7, 8, 9]

> [op(4.., a)];
                              [4, 5, 6, 7, 8, 9]

saves you from typing

> [op(4..nops(a), a)];
                              [4, 5, 6, 7, 8, 9]

We can apply the same trick for suffices, but this doesn’t save as much typing.

> [op(..5, a)];
                                [1, 2, 3, 4, 5]

Sadly, this convention is not applied uniformly,

> sum(x^i,i=1..infinity);
                                         x
                                     - -----
                                       x - 1
> sum(x^i,i=1..);
Error, (in sum) cannot split rhs for multiple assignment

and again, the convention is unusable if you care about backwards compatibility.

$\|\hat{f}(x)-\hat{f}(y)\|$	$=\| \inf_{a \in A} \{f(a)+d(a,x)\} – \inf_{b \in A} \{f(b)+d(b,y)\}\|$
	$\leq \| \inf_{a \in A} \{f(a)+d(a,x)\} – (f(b_0)+d(b_0,y))\|$
	$\leq \| f(b_0)+d(b_0,x) – f(b_0)-d(b_0,y)\|$
	$\leq \|d(b_0,x) – d(b_0,y)\|$
	$\leq d(x,y)$
	$\leq d(x,a) + d(a,y) + f(a) + f(a), (\forall a \in A)$