Jones’ theorem on the cardinal invariant $\mathfrak{p}$

This post continues the study of the cardinal invariant $\mathfrak{p}$. We refer the reader to a previous post for all the needed background. For ordinals $\alpha ,{\alpha }_0,{\alpha }_1,\beta ,{\beta }_0,{\beta }_1$, the polarized partition relation $$\left(\begin{array}{c}\alpha \\ \beta \end{array}\right)\to \left(\begin{array}{cc}{\alpha }_0& {\alpha }_1\\ {\beta }_0& {\beta }_1\end{array}\right)$$ asserts that for every coloring $f:\alpha \times \beta \to 2$, (at least) one of the following holds:

1. there are $A\subseteq \alpha$ and $B\subseteq \beta$ with $\text{otp}(A)={\alpha }_0,\text{otp}(B)={\beta }_0$ s.t. $f[A\times B]=\{0\}$;
2. there are $A\subseteq \alpha$ and $B\subseteq \beta$ with $\text{otp}(A)={\alpha }_1,\text{otp}(B)={\beta }_1$ s.t. $f[A\times B]=\{1\}$.

Erdős and Rado proved that $$\left(\begin{array}{c}{\omega }_1\\ \omega \end{array}\right)\to \left(\begin{array}{cc}{\omega }_1& n\\ \omega & \omega \end{array}\right)$$ holds for all $n<\omega$. The current post will center around a proof of the following relative:

Theorem (A.L. Jones, 2008). For all $\alpha <\mathfrak{p}$: $$\left(\begin{array}{c}\mathfrak{p}\\ \omega \end{array}\right)\to \left(\begin{array}{cc}\mathfrak{p}& \alpha \\ \omega & \omega \end{array}\right).$$

Proof. Suppose that we are given a coloring $f:\mathfrak{p}\times \omega \to 2$. Denote $$1_{\tau }:=\{n<\omega \mid f(\tau ,n)=1\},$$ and let $X$ be a maximal (with respect to inclusion) subset of $\mathfrak{p}$ with the property that the intersection of any finite collection of subsets from ${\mathcal{A}}_X:=\{1_{\tau }\mid \tau \in X\}$ is infinite. The proof is now divided into two.

$▸$ Suppose that $|X|<\mathfrak{p}$.
By $|{\mathcal{A}}_X|<\mathfrak{p}$, let us fix an infinite pseudointersection $P$ for ${\mathcal{A}}_X$ (indeed, if $X=\mathrm{\varnothing }$, we simply take $P:=\omega$). Notice that if $Q\subseteq \omega$ and $P\cap Q$ is infinite, then $Q\cap \bigcap _{\tau \in I}{1}_{\tau }$ is infinite for all $I\in [X{]}^{<\omega }$. Thus, it follows from the maximality of $X$ that $P\cap 1_{\tau }$ is finite for all $\tau \in \mathfrak{p}\setminus X$. Put $Y:=\mathfrak{p}\setminus X$ and define $g:Y\to [P{]}^{<\omega }$ by letting $g(\tau ):=P\cap 1_{\tau }$ for all $\tau \in Y$. Then, by $\mathfrak{p}=\text{cf}(\mathfrak{p})>\omega$, we may find a subset $A\subseteq Y$ of size $\mathfrak{p}$ on which $g$ is constant. In particular, $B:=\bigcap _{\tau \in A}(\omega \setminus 1_{\tau })$ contains the infinite set $P\setminus g(min(A))$.
Altogether, $\text{otp}(A)=\mathfrak{p}$, $\text{otp}(B)=\omega$, and $f[A\times B]=\{0\}$.

$▸$ Suppose that $|X|=\mathfrak{p}$.
Let $⟨M_{\alpha }\mid \alpha <\mathfrak{p}⟩$ be a continuous $ϵ$-chain of elementary submodels of $(\mathcal{H}(\theta ),\in ,{<}_{\theta })$ for a large enough regular $\theta$, such that $f,X\in M_0$ and $M_{\alpha }\cap \mathfrak{p}\in \mathfrak{p}$ for all $\alpha <\mathfrak{p}$.
Suppose that we are given a limit ordinal $\alpha <\mathfrak{p}$. Pick $\chi \in (X\cap M_{\alpha +1})\setminus M_{\alpha }$, and consider the poset $\mathbb{P}:=⟨P,\le ⟩$, where $$P:=\{(a,b)\in [X\cap M_{\alpha }{]}^{<\omega }\times [\omega {]}^{<\omega }\mid \chi \in a,f[a\times b]=\{1\}\},$$ and $(a_0,b_0)\le (a_1,b_1)$ iff $a_0\subseteq a_1$ and $b_0\subseteq b_1$.
For all $b\in [\omega {]}^{<\omega }$, put $P_b:=\{(a^{\prime },b^{\prime })\in P\mid b^{\prime }=b\}$. Then $⟨P_b\mid b\in [\omega {]}^{<\omega }⟩$ witnesses that $\mathbb{P}$ is $\sigma$-centered.

For all $n<\omega$, let $$D_n:=\{(a,b)\in P\mid max(b)>n\}.$$ Recalling that the intersection of any finite collection of subsets from $\{1_{\tau }\mid \tau \in X\}$ is infinite, we see that $D_n$ is cofinal in $\mathbb{P}$.

For all $\beta <\alpha$, let $$E_{\beta }:=\{(a,b)\in P\mid (a\cap M_{\beta +1})\setminus M_{\beta }\ne \mathrm{\varnothing }\}.$$Then, by $\chi \in (a\cap M_{\alpha +1})\setminus M_{\alpha }$ for all $(a,b)\in P$, and a standard elementarity argument, we see that $E_{\beta }$ is cofinal in $\mathbb{P}$.

Finally, let us pick some subset $G(\alpha )\subseteq P$ such that:

1. $\{p{\}}^{\uparrow }\cap \{q{\}}^{\uparrow }\ne \mathrm{\varnothing }$ for all $p,q\in G(\alpha )$;
2. $G(\alpha {)}_{\downarrow }\cap D_n\ne \mathrm{\varnothing }$ for all $n<\omega$;
3. $G(\alpha {)}_{\downarrow }\cap E_{\beta }\ne \mathrm{\varnothing }$ for all $\beta <\alpha$.

Put $A:=\bigcup \{a\mid (a,b)\in G(\alpha )\}$ and $B:=\bigcup \{b\mid (a,b)\in G(\alpha )\}$.
Then item 1 entails that $f[a\times b]=\{1\}$, item 2 entails that $\text{otp}(B)=\omega$, and item 3 entails that $\text{otp}(A)\ge \alpha$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Let us point out that the above proof works equally well for stationary sets.

Proposition. For every stationary subset $S$ of a regular uncountable cardinal $\kappa$, and every coloring $f:S\times \omega \to 2$, one of the following holds:

1. there exists a stationary $A\subseteq S$ and an infinite $B\subseteq \omega$ such that $f[A\times B]=\{0\}$;
2. for every $ϵ<min\{\kappa ,\mathfrak{p}\}$, there exists $A\subseteq S$ of order-type $ϵ$ and an infinite $B\subseteq \omega$ such that $f[A\times B]=\{1\}$.

Proof (sketch). Given the coloring $f:S\times \omega \to 2$, let $1_{\tau }:=\{n<\omega \mid f(\tau ,n)=1\}$. Let $X$ be a maximal subset of $S$ with the property that the intersection of any finite collection of subsets from ${\mathcal{A}}_X:=\{1_{\tau }\mid \tau \in X\}$ is infinite.
Write $\mu :=min\{\kappa ,\mathfrak{p}\}$, and consider two complimentary cases.

$▸$ Suppose that $|X|<\mu$.
Fix an infinite pseudointersection $P$ for ${\mathcal{A}}_X$ (or let $P=\omega$ in case that $X$ is empty). Then $P\cap 1_{\tau }$ is finite for all $\tau \in S\setminus X$. Define $g:S\setminus X\to [P{]}^{<\omega }$ by stipulating that $g(\tau ):=P\cap 1_{\tau }$. Then, by $\text{cf}(\kappa )>\omega$, we may find a stationary $A\subseteq S\setminus X$ on which $g$ is constant. Put $B:=\bigcap _{\tau \in A}(\omega \setminus 1_{\tau })$. Then $B$ is infinite, and $f[A\times B]=\{0\}$.

$▸$ Suppose that $|X|\ge \mu$.
Fix a limit $ϵ<\mu$. Let $⟨M_{\alpha }\mid \alpha <ϵ+1⟩$ be a continuous $ϵ$-chain of elementary submodels with $f,X\in M_0$ and $M_{\alpha }\cap \mu \in \mu$ for all $\alpha$. Pick $\chi \in X\setminus M_{ϵ}$, and consider the poset $\mathbb{P}:=⟨P,\le ⟩$, where $P:=\{(a,b)\in [X\cap M_{ϵ}{]}^{<\omega }\times [\omega {]}^{<\omega }\mid \chi \in a,f[a\times b]=\{1\}\}$, and $(a_0,b_0)\le (a_1,b_1)$ iff $a_0\subseteq a_1$ and $b_0\subseteq b_1$. As before, $\mathbb{P}$ is $\sigma$-centered. Also, for all $n<\omega$ and $\beta <ϵ$, the set $$D_n^{\beta }:=\{(a,b)\in P\mid (a\cap M_{\beta +1})\setminus M_{\beta }\ne \mathrm{\varnothing }\text{ and }|b|>n\}$$ is cofinal in $\mathbb{P}$. Thus, by $ϵ<\mathfrak{p}$, we can pick a subset $G\subseteq P$ such that $\{p{\}}^{\uparrow }\cap \{q{\}}^{\uparrow }\ne \mathrm{\varnothing }$ for all $p,q\in G$, and $G_{\downarrow }\cap D_n^{\beta }\ne \mathrm{\varnothing }$ for all $n<\omega$ and $\beta <ϵ$.

Put $A:=\bigcup \{a\mid (a,b)\in G(\alpha )\}$ and $B:=\bigcup \{b\mid (a,b)\in G(\alpha )\}$. Then $B$ is infinite, $\text{otp}(A)\ge ϵ$, and $f[a\times b]=\{1\}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

As $\mathfrak{p}$ is provably uncountable in ZFC, Jones’ theorem shows that the integer $n$ in the above-cited theorem of Erdos and Rado, may be replaced with any countable ordinal $\alpha$. We conclude this post by addressing the remaining case, that is, the case of ${\omega }_1$.

Theorem (Sierpiński). CH entails $$\left(\begin{array}{c}{\omega }_1\\ \omega \end{array}\right)\nrightarrow \left(\begin{array}{cc}{\omega }_1& {\omega }_1\\ \omega & \omega \end{array}\right).$$

Proof. Sierpinski proved in the 1930’s that CH implies the existence of sequence of functions $⟨f_n:\mathbb{R}\to \mathbb{R}\mid n<\omega ⟩$ such that for every uncountable set of reals $A$, there exists some $m<\omega$ such that $f_n[A]=\mathbb{R}$ for all $n\ge m$. By CH, let us also fix a bijection $\psi :{\omega }_1\leftrightarrow \mathbb{R}$. Finally, define $f:{\omega }_1\times \omega \to 2$ by letting $f(\tau ,n):=0$ iff $f_n(\psi (\tau ))=0$.
Then $f$ witnesses the failure of $\left(\begin{array}{c}{\omega }_1\\ \omega \end{array}\right)\to \left(\begin{array}{cc}{\omega }_1& {\omega }_1\\ \omega & \omega \end{array}\right)$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

In contrast:

Theorem (Laver, 1973). $M{A}_{{\mathrm{\aleph }}_1}$ entails$$\left(\begin{array}{c}{\omega }_1\\ \omega \end{array}\right)\to \left(\begin{array}{cc}{\omega }_1& {\omega }_1\\ \omega & \omega \end{array}\right).$$

Moreover, for every ordinal $\kappa <\mathfrak{p}$ of uncountable cofinality, any nonzero limit ordinal $\alpha <{\omega }_1$, and any positive integer $n<\omega$ we have:$$\left(\begin{array}{c}\kappa \\ {\omega }^{\alpha }\end{array}\right)\to {\left(\begin{array}{c}\kappa \\ {\omega }^{\alpha }\end{array}\right)}_n.$$
Proof. Given a coloring $f:\kappa \times {\omega }^{\alpha }\to n$, write $A_{i,\zeta }:=\{\beta <{\omega }^{\alpha }\mid f(\beta ,\zeta )=i\}$. Then by Lemma 4 from a previous post, there exists a function $g:\kappa \to n$ and a set $B\subseteq {\omega }^{\alpha }$ of order-type ${\omega }^{\alpha }$ such that $B{\subseteq }^{\ast }{A}_{g(\zeta ),\zeta }$ for all $\zeta <\kappa$. Fix $i^{\ast }<n$ such that $g^{-1}\{i^{\ast }\}$ is cofinal in $\kappa$, and put $A:=g^{-1}\{i^{\ast }\}$. Let $\{{\beta }_k\mid k<\omega \}$ be some cofinal subset of ${\omega }^{\alpha }$, and then, define $h:A\to \omega$ by stipulating $$h(\zeta ):=min\{k<\omega \mid (B\setminus {\beta }_k)\subseteq A_{i^{\ast },\zeta }\}.$$As $\text{cf}(\kappa )>\omega$, let us fix $k^{\ast }<\omega$ such that $h^{-1}\{k^{\ast }\}$ is cofinal in $\kappa$.  Put $A^{\ast }:=h^{-1}\{k^{\ast }\}$ and $B^{\ast }:=B\setminus {\beta }_{k^{\ast }}$. Then $\text{otp}(A^{\ast })=\kappa ,\text{otp}(B^{\ast })={\omega }^{\alpha }$, and $f[A\times B]=\{i^{\ast }\}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$