The uniformization property for ${\mathrm{\aleph }}_2$

Given a subset of a regular uncountable cardinal $S\subseteq \kappa$, $U{P}_S$ (read: “the uniformization property holds for $S$”) asserts that for every sequence $\overrightarrow{f}=⟨f_{\alpha }\mid \alpha \in S⟩$ satisfying for all $\alpha \in S$:

1. $f_{\alpha }$ is a 2-valued function;
2. $\text{dom}(f_{\alpha })$ is a cofinal subset of $\alpha$ of minimal order-type,

there exists a function $f:\kappa \to 2$ such that for every limit $\alpha \in S$, the set $\{\beta \in \text{dom}(f_{\alpha })\mid f_{\alpha }(\beta )\ne f(\beta )\}$ is bounded in $\alpha$. Idea: we think of $f$ as a coloring “uniformizing” the sequence of local colorings $\overrightarrow{f}$.

In a previous post, we proved that $M{A}_{{\mathrm{\aleph }}_1}$ entails $U{P}_{E_{\omega }^{{\omega }_1}}$. The same proof shows that $M{A}_{{\mathrm{\aleph }}_2}$ entails $U{P}_{E_{\omega }^{{\omega }_2}}$. In this post, we shall focus on $U{P}_S$ for $S=E_{{\omega }_1}^{{\omega }_2}$. More specifically, we shall provide a proof to Shelah’s ZFC theorem that $U{P}_{E_{{\omega }_1}^{{\omega }_2}}$ fails.

We commence with a simple observation.

Observation. Suppose that $U{P}_S$ holds for a given stationary subset $S\subseteq E_{{\omega }_1}^{{\omega }_2}$. Then
(a) CH holds, and
(b) for every sequence $\overrightarrow{f}=⟨f_{\alpha }\mid \alpha \in S⟩$ satisfying for all $\alpha \in S$:

1. $f_{\alpha }$ is an ${\omega }_1$-valued function;
2. $\text{dom}(f_{\alpha })$ is a cofinal subset of $\alpha$ of order-type ${\omega }_1$,

there exists a function $f:{\omega }_2\to {\omega }_1$ such that for every $\alpha \in S$, the set $\{\beta \in \text{dom}(f_{\alpha })\mid f_{\alpha }(\beta )\ne f(\beta )\}$ is bounded in $\alpha$.
Proof. Suppose that we are given a sequence of local colorings $\overrightarrow{f}=⟨f_{\alpha }:A_{\alpha }\to {}^{w}2\mid \alpha \in S⟩$ with each $A_{\alpha }$ being a cofinal subset of $\alpha$ of order-type ${\omega }_1$. Since $U{P}_S$ holds, we get that for every $n<\omega$, there exists a function $f^n:{\omega }_2\to 2$ such that for every $\alpha \in S$, the set $\{\beta \in A_{\alpha }\mid f_{\alpha }(\beta )(n)\ne f^n(\beta )\}$ is bounded in $\alpha$. Define $f:{\omega }_2\to {}^{\omega }2$ by stipulating that $f(\beta )(n)=f^n(\beta )$. Then, for every $\alpha \in S$, we have $\text{cf}(\alpha )>\omega$, and hence the set $\{\beta \in A_{\alpha }\mid f_{\alpha }(\beta )\ne f(\beta )\}$ is bounded in $\alpha$.

Since $|{}^{\omega }2|\ge {\mathrm{\aleph }}_1$, the above establishes (b).
Now, to see that (a) holds, assume towards a contradiction that CH fails. In particular, $|{}^{\omega }2|\ge {\mathrm{\aleph }}_2$, and the sequence of constant functions $\overrightarrow{f}:=⟨f_{\alpha }:A_{\alpha }\to \{\alpha \}\mid \alpha \in S⟩$ must have a uniformization, $f:{\omega }_2\to {\omega }_2$. As $S$ is stationary, let us pick some $\alpha \in S$ such that $f[\alpha ]\subseteq \alpha$. Then $\{\beta \in A_{\alpha }\mid f_{\alpha }(\beta )\ne f(\beta )\}=A_{\alpha }$, contradicting the assumption that $f$ is a uniformizing function for $\overrightarrow{f}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Main Theorem (Shelah). $U{P}_{E_{{\omega }_1}^{{\omega }_2}}$ fails.
Proof.  Suppose not. Then by the above observation, CH holds. In particular, $|H({\omega }_1)|={\mathrm{\aleph }}_1$. Denote $S:=E_{{\omega }_1}^{{\omega }_2}$. For every $\delta \in S$, pick an increasing and continuous function $c_{\delta }:{\omega }_1\to \delta$ whose range is cofinal in $\delta$. It then follows from the previous observation that for every sequence $\overrightarrow{f}=⟨f_{\delta }:{\omega }_1\to H({\omega }_1)\mid \delta \in S⟩$, there exists a function $f:{\omega }_2\to H({\omega }_1)$ together with a function $r:S\to {\omega }_1$ such that for every $\delta \in S$ we have $f(c_{\delta }(j))=f_{\delta }(j)$ whenever $r(\delta )\le j<{\omega }_1$.

Fix a large enough regular cardinal $\lambda >2^{{\mathrm{\aleph }}_2}$, and put $\mathfrak{A}:=(H(\lambda ),\in ,{<}_{\lambda })$, where ${<}_{\lambda }$ is some well-ordering of $H(\lambda )$. For every $\delta \in S$, $i<{\omega }_1$, and $p\in H(\lambda )$, let $N_{\delta ,i}^p$ denote the Skolem hull of $\{p,\delta ,i\}$ in $\mathfrak{A}$.
Let ${\pi }_{\delta ,i}^p:N_{\delta ,i}^p\leftrightarrow M_{\delta ,i}^p$ denote the  collapsing function from $(N_{\delta ,i}^p,\in )$ to a transitive model $(M_{\delta ,i}^p,\in )$.  Put:
$${<}_{\delta ,i}^p:=\{(x,y)\in M_{\delta ,i}^p\times M_{\delta ,i}^p\mid ({\pi }_{\delta ,i}^p{)}^{-1}(x){<}_{\lambda }({\pi }_{\delta ,i}^p{)}^{-1}(y)\}.$$

By CH and a result from a previous post, we may fix a function $G:[{\omega }_2{]}^{<{\omega }_1}\to H({\omega }_1)$ with the property that $G(A)=G(B)$ entails the existence of an order-preserving isomorphism $g:A\leftrightarrow B$ which is the identity on $A\cap B$.

Next, we define a sequence $⟨(p_n,⟨f_{\delta }^n\mid \delta \in S⟩,h_n,r_n)\mid n<\omega ⟩$ by recursion, as follows.

• let $p_0:=⟨⟨c_{\delta }\mid \delta \in S⟩⟩$;
• let $p_{n+1}:=⟨p_n,h_n,r_n⟩$;
• let $⟨f_{\delta }^n:{\omega }_1\to H({\omega }_1)\mid \delta \in S⟩$ be such that for all $i<{\omega }_1$:
  $$f_{\delta }^n(i)=⟨M_{\delta ,i}^{p_n},{\pi }_{\delta ,i}^{p_n}(p_n),{\pi }_{\delta ,i}^{p_n}(\delta ),{\pi }_{\delta ,i}^{p_n}(i),{<}_{\delta ,i}^{p_n},G(N_{\delta ,i}^{p_n}\cap {\omega }_2)⟩.$$
• let $h_n:{\omega }_2\to H({\omega }_1)$ and $r_n:{\omega }_2\to {\omega }_1$ be such that $h_n(c_{\delta }(j))=f_{\delta }^n(j)$ whenever $r_n(\delta )\le j<{\omega }_1$.

Define $r:S\to {\omega }_1$, by stipulating that $r(\delta )=\underset{n<\omega }{sup}{r}_n(\delta )$.

Subclaim. There exists $\delta \in S$ and $j<{\omega }_1$ for which the following set is non-empty:
$$T_j^{\delta }:=\{\gamma \in S\mid \gamma >\delta ,j\ge max\{r(\delta ),r(\gamma )\},c_{\gamma }\upharpoonright (j+1)=c_{\delta }\upharpoonright (j+1)\}.$$
Proof. Fix $j<{\omega }_1$, such that $T_j:=\{\delta \in S\mid r(\delta )=j\}$ is stationary. Then, by CH, the set $\{c_{\delta }\upharpoonright (j+1)\mid \delta \in T_j\}$ has size ${\omega }_1$, and hence we can find ${\delta }_1<{\delta }_2$ in $T_j$ such that $c_{{\delta }_1}\upharpoonright j+1=c_{{\delta }_2}\upharpoonright j+1$. Clearly, ${\delta }_2\in T_j^{{\delta }_1}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$

Let ${\delta }_1:=min\{\delta \in S\mid \mathrm{\exists }j<{\omega }_1(T_j^{\delta }\ne \mathrm{\varnothing })\}$, and then ${\delta }_2:=min\bigcup \{T_j^{{\delta }_1}\mid j<{\omega }_1\}$. Put $i:=min\{i^{\prime }<{\omega }_1\mid c_{{\delta }_1}(i^{\prime })\ne c_{{\delta }_2}(i^{\prime })\}$. Since $c_{{\delta }_1}$ and $c_{{\delta }_2}$ are continuous, $i$ is a successor ordinal. Denote $j:=i-1$. Then $c_{{\delta }_1}\upharpoonright j+1=c_{{\delta }_2}\upharpoonright j+1$, and hence $j\ge max\{r({\delta }_1),r({\delta }_2)\}$.

For all $n<\omega$, by definition of $p_{n+1}$, we have $p_n\in N_{{\delta }_1,j}^{p_{n+1}}$ and hence $N_{{\delta }_1,j}^{p_n}\in N_{{\delta }_1,j}^{p_{n+1}}$.
So $N_1:=\bigcup _{n<\omega }{N}_{{\delta }_1,j}^{p_n}$, and $N_2:=\bigcup _{n<\omega }{N}_{{\delta }_2,j}^{p_n}$, are elementary submodels of $(H(\lambda ),\in ,{<}_{\lambda })$.

Subclaim. There exists an isomorphism $g:N_1\to N_2$ such that $g\upharpoonright (N_1\cap N_2\cap {\omega }_2)$ is the identity function, $g({\delta }_1)={\delta }_2$, and for all $n<\omega$: $g\upharpoonright N_{{\delta }_1,j}^{p_n}$ is an isomorphism from $N_{{\delta }_1,j}^{p_n}$ to $N_{{\delta }_2,j}^{p_n}$.
Proof. By $j\ge max\{r({\delta }_1),r({\delta }_2)\}$ and ${\delta }_2\in T_j^{{\delta }_1}$, we have for every $n<\omega$, $$f_{{\delta }_1}^n(j)=h_n(c_{{\delta }_1}(j))=h_n(c_{{\delta }_2}(j))=f_{{\delta }_2}^n(j),$$ and hence
$$\begin{array}{c}⟨M_{{\delta }_1,j}^{p_n},{\pi }_{{\delta }_1,j}^{p_n}(p_n),{\pi }_{{\delta }_1,j}^{p_n}({\delta }_1),{\pi }_{{\delta }_1,j}^{p_n}(j),{<}_{{\delta }_1,j}^{p_n},G(N_{{\delta }_1,j}^{p_n}\cap {\omega }_2)⟩=\\ ⟨M_{{\delta }_2,j}^{p_n},{\pi }_{{\delta }_2,j}^{p_n}(p_n),{\pi }_{{\delta }_2,j}^{p_n}({\delta }_2),{\pi }_{{\delta }_2,j}^{p_n}(j),{<}_{{\delta }_2,j}^{p_n},G(N_{{\delta }_2,j}^{p_n}\cap {\omega }_2)⟩\end{array}.$$

So $g_n:=({\pi }_{{\delta }_2,j}^{p_n}{)}^{-1}\circ {\pi }_{{\delta }_1,j}^{p_n}:N_{{\delta }_1,j}^{p_n}\to N_{{\delta }_2,j}^{p_n}$ is an isomorphism that satisfies:

• $g_n(p_n)=p_n$;
• $g_n({\delta }_1)={\delta }_2$;
• $g_n(j)=j$;
• $x{<}_{\lambda }y$ iff $g_n(x){<}_{\lambda }{g}_n(y)$ for all $x,y\in N_{{\delta }_1,j}^{p_n}$.

As well-ordering is a rigid relation, the last property implies that $g_n$ is unique (though, we could have used the $\in$ relation for that). In particular, $g_n\subseteq g_{n+1}$, and then the fact that  $G(N_{{\delta }_1,j}^{p_n}\cap {\omega }_2)=G(N_{{\delta }_2,j}^{p_n}\cap {\omega }_2)$ implies that $g_n\upharpoonright (N_{{\delta }_1,j}^{p_n}\cap N_{{\delta }_2,j}^{p_n}\cap {\omega }_2)$ is the identity function.
Let $g:=\bigcup _{n<\omega }{g}_n$. Then $g$ is an isomorphism from $N_1$ to $N_2$, and $g\upharpoonright (N_1\cap N_2\cap {\omega }_2)$ is the identity function. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$

Let  ${\gamma }_1:=min(N_1\cap {\omega }_2\setminus (N_1\cap N_2))$, and ${\gamma }_2:=min(N_2\cap {\omega }_2\setminus (N_1\cap N_2))$.

Evidently, ${\gamma }_1\ne {\gamma }_2$ and $g({\gamma }_1)={\gamma }_2$. Note that $\text{cf}({\gamma }_1)=\text{cf}({\gamma }_2)={\omega }_1$, because otherwise, $\text{cf}({\gamma }_1)=\text{cf}({\gamma }_2)=\omega$, in which case, for the least $n<\omega$ such that ${\gamma }_1\in N_{{\delta }_1,j}^{p_n}$, there would have existed a countable set $A\subseteq {\gamma }_1\cap N_{{\delta }_1,j}^{p_n}$ which is cofinal in ${\gamma }_1$. Recalling that $g\upharpoonright N_1\cap {\gamma }_1$ is the identity function, we would have then get that $A\subseteq N_{{\delta }_2,j}^{p_n}$, and then ${\gamma }_1$ would have been an element of $N_{{\delta }_2,j}^{p_{n+1}}$ (note that ${\gamma }_2\in N_{{\delta }_2,j}^{p_n}$ and $sup(N_{{\delta }_2,j}^{p_n}\cap {\gamma }_2)={\gamma }_1$), contradicting the fact that ${\gamma }_1\notin N_2$.

So ${\gamma }_1$ and ${\gamma }_2$ are distinct elements of $S$. Since $g({\delta }_1)={\delta }_2\ne {\delta }_1$, we get that ${\delta }_1\notin N_2$ and ${\delta }_2\notin N_1$. Consequently, ${\gamma }_1\le {\delta }_1$ and ${\gamma }_2\le {\delta }_2$.

Denote $l:=N_1\cap {\omega }_1$. As $N_1$ and $N_2$ are countable isomorphic models, we have $g({\omega }_1)={\omega }_1$, and $N_2\cap {\omega }_1=l$.
Since $g({\gamma }_1)={\gamma }_2$ and $g\upharpoonright N_1\cap {\gamma }_1$ is the identity function, we get  for all $t<l$ that $c_{{\gamma }_1}(t)=g(c_{{\gamma }_1}(t))=c_{{\gamma }_2}(g(t))=c_{{\gamma }_2}(t)$. So, continuity now tells us that $c_{{\gamma }_1}(l)=c_{{\gamma }_2}(l)$. In particular, ${\gamma }_2\in T_l^{{\gamma }_1}$. By the minimality condition in the definitions of  ${\delta }_1,{\delta }_2$, we then infer that ${\gamma }_1={\delta }_1$ and ${\gamma }_2={\delta }_2$. Altogether, we got that $c_{{\delta }_1}\upharpoonright l+1=c_{{\delta }_2}\upharpoonright l+1$, and hence $l\le j$. On the other hand, $j\in N_{{\delta }_1,j}^{p_0}$, and hence $j<N_1\cap {\omega }_2=l$. This is a contradiction. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$