Given a subset of a regular uncountable cardinal $S\subseteq\kappa$, $UP_S$ (read: “the uniformization property holds for $S$”) asserts that for every sequence $\overrightarrow f=\langle f_\alpha\mid \alpha\in S\rangle$ satisfying for all $\alpha\in S$:
- $f_\alpha$ is a 2-valued function;
- $\text{dom}(f_\alpha)$ is a cofinal subset of $\alpha$ of minimal order-type,
there exists a function $f:\kappa\rightarrow 2$ such that for every limit $\alpha\in S$, the set $\{ \beta\in\text{dom}(f_\alpha)\mid f_\alpha(\beta)\neq f(\beta)\}$ 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 $MA_{\aleph_1}$ entails $UP_{E^{\omega_1}_{\omega}}$. The same proof shows that $MA_{\aleph_2}$ entails $UP_{E^{\omega_2}_{\omega}}$. In this post, we shall focus on $UP_S$ for $S=E^{\omega_2}_{\omega_1}$. More specifically, we shall provide a proof to Shelah’s ZFC theorem that $UP_{E^{\omega_2}_{\omega_1}}$ fails.
We commence with a simple observation.
Observation. Suppose that $UP_S$ holds for a given stationary subset $S\subseteq E^{\omega_2}_{\omega_1}$. Then
(a) CH holds, and
(b) for every sequence $\overrightarrow f=\langle f_\alpha\mid \alpha\in S\rangle$ satisfying for all $\alpha\in S$:
- $f_\alpha$ is an $\omega_1$-valued function;
- $\text{dom}(f_\alpha)$ is a cofinal subset of $\alpha$ of order-type $\omega_1$,
there exists a function $f:\omega_2\rightarrow\omega_1$ such that for every $\alpha\in S$, the set $\{ \beta\in\text{dom}(f_\alpha)\mid f_\alpha(\beta)\neq f(\beta)\}$ is bounded in $\alpha$.
Proof. Suppose that we are given a sequence of local colorings $\overrightarrow f=\langle f_\alpha:A_\alpha\rightarrow{}^w2\mid \alpha\in S\rangle$ with each $A_\alpha$ being a cofinal subset of $\alpha$ of order-type $\omega_1$. Since $UP_S$ holds, we get that for every $n<\omega$, there exists a function $f^n:\omega_2\rightarrow 2$ such that for every $\alpha\in S$, the set $\{ \beta\in A_\alpha\mid f_\alpha(\beta)(n)\neq f^n(\beta)\}$ is bounded in $\alpha$. Define $f:\omega_2\rightarrow{}^\omega2$ 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 $\{ \beta\in A_\alpha\mid f_\alpha(\beta)\neq f(\beta)\}$ is bounded in $\alpha$.
Since $|{}^\omega2|\ge\aleph_1$, the above establishes (b).
Now, to see that (a) holds, assume towards a contradiction that CH fails. In particular, $|{}^\omega2|\ge\aleph_2$, and the sequence of constant functions $\overrightarrow f:=\langle f_\alpha:A_\alpha\rightarrow\{\alpha\}\mid \alpha\in S\rangle$ must have a uniformization, $f:\omega_2\rightarrow\omega_2$. As $S$ is stationary, let us pick some $\alpha\in S$ such that $f[\alpha]\subseteq\alpha$. Then $\{\beta\in A_\alpha\mid f_\alpha(\beta)\neq f(\beta)\}=A_\alpha$, contradicting the assumption that $f$ is a uniformizing function for $\overrightarrow f$. $\square$
Main Theorem (Shelah). $UP_{E^{\omega_2}_{\omega_1}}$ fails.
Proof. Suppose not. Then by the above observation, CH holds. In particular, $|H(\omega_1)|=\aleph_1$. Denote $S:=E^{\omega_2}_{\omega_1}$. For every $\delta\in S$, pick an increasing and continuous function $c_\delta:\omega_1\rightarrow\delta$ whose range is cofinal in $\delta$. It then follows from the previous observation that for every sequence $\overrightarrow f=\langle f_\delta:\omega_1\rightarrow H(\omega_1)\mid \delta\in S\rangle$, there exists a function $f:\omega_2\rightarrow H(\omega_1)$ together with a function $r:S\rightarrow\omega_1$ 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 $\lambda>2^{\aleph_2}$, 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 $N^p_{\delta,i}$ denote the Skolem hull of $\{p,\delta,i\}$ in $\mathfrak A$.
Let $\pi^p_{\delta,i}:N^p_{\delta,i}\leftrightarrow M^p_{\delta,i}$ denote the collapsing function from $(N^p_{\delta,i},\in)$ to a transitive model $(M^p_{\delta,i},\in)$. Put:
$$<^p_{\delta,i}:=\{ (x,y)\in M^{p}_{\delta,i}\times M^{p}_{\delta,i}\mid (\pi^{p}_{\delta,i})^{-1}(x)<_\lambda (\pi^{p}_{\delta,i})^{-1}(y)\}.$$
By CH and a result from a previous post, we may fix a function $G:[\omega_2]^{<\omega_1}\rightarrow H(\omega_1)$ 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 $\langle (p_n,
\langle f_\delta^n\mid \delta\in S \rangle, h_n,r_n)\mid n<\omega\rangle$ by recursion, as follows.
- let $p_0:=\langle \langle c_\delta\mid \delta\in S\rangle\rangle$;
- let $p_{n+1}:=\langle p_n,h_n,r_n\rangle$;
- let $\langle f^n_\delta:\omega_1\rightarrow H(\omega_1)\mid \delta\in S\rangle$ be such that for all $i<\omega_1$:
$$f^n_\delta(i)=\langle M^{p_n}_{\delta,i},\pi^{p_n}_{\delta,i}(p_n),\pi^{p_n}_{\delta,i}(\delta),\pi^{p_n}_{\delta,i}(i),<^{p_n}_{\delta,i}, G(N^{p_n}_{\delta,i}\cap\omega_2)\rangle.$$ - let $h_n:\omega_2\rightarrow H(\omega_1)$ and $r_n:\omega_2\rightarrow\omega_1$ be such that $h_n(c_\delta(j))=f^n_\delta(j)$ whenever $r_n(\delta)\le j<\omega_1$.
Define $r:S\rightarrow\omega_1$, by stipulating that $r(\delta)=\sup_{n<\omega}r_n(\delta)$.
Subclaim. There exists $\delta\in S$ and $j<\omega_1$ for which the following set is non-empty:
$$T^{\delta}_j:=\{\gamma\in S \mid \gamma>\delta, j\ge\max\{r(\delta),r(\gamma)\}, c_{\gamma}\restriction(j+1)=c_{\delta}\restriction(j+1)\}.$$
Proof. Fix $j<\omega_1$, such that $T_j:= \{\delta\in S\mid r(\delta)=j\}$ is stationary. Then, by CH, the set $\{ c_\delta\restriction(j+1)\mid \delta\in T_j\}$ has size $\omega_1$, and hence we can find $\delta_1<\delta_2$ in $T_j$ such that $c_{\delta_1}\restriction j+1=c_{\delta_2}\restriction j+1$. Clearly, $\delta_2\in T^{\delta_1}_j$. $\blacksquare$
Let $\delta_1:=\min\{\delta\in S\mid \exists j<\omega_1(T^\delta_j\neq\emptyset)\}$, and then $\delta_2:=\min\bigcup\{ T^{\delta_1}_j\mid j<\omega_1\}$. Put $i:=\min\{ i'<\omega_1\mid c_{\delta_1}(i’)\neq c_{\delta_2}(i’)\}$. Since $c_{\delta_1}$ and $c_{\delta_2}$ are continuous, $i$ is a successor ordinal. Denote $j:=i-1$. Then $c_{\delta_1}\restriction j+1=c_{\delta_2}\restriction j+1$, and hence $j\ge\max\{r(\delta_1),r(\delta_2)\}$.
For all $n<\omega$, by definition of $p_{n+1}$, we have $p_n\in N^{p_{n+1}}_{\delta_1,j}$ and hence $N^{p_n}_{\delta_1,j}\in N^{p_{n+1}}_{\delta_1,j}$.
So $N_1:=\bigcup_{n<\omega}N^{p_{n}}_{\delta_1,j}$, and $N_2:=\bigcup_{n<\omega}N^{p_{n}}_{\delta_2,j}$, are elementary submodels of $(H(\lambda),\in,<_\lambda)$.
Subclaim. There exists an isomorphism $g:N_1\rightarrow N_2$ such that $g\restriction(N_1\cap N_2\cap\omega_2)$ is the identity function, $g(\delta_1)=\delta_2$, and for all $n<\omega$: $g\restriction N^{p_n}_{\delta_1,j}$ is an isomorphism from $N^{p_n}_{\delta_1,j}$ to $N^{p_n}_{\delta_2,j}$.
Proof. By $j\ge\max\{r(\delta_1),r(\delta_2)\}$ and $\delta_2\in T^{\delta_1}_j$, we have for every $n<\omega$, $$f^n_{\delta_1}(j)=h_n(c_{\delta_1}(j))=h_n(c_{\delta_2}(j))=f^n_{\delta_2}(j),$$ and hence
$$\begin{array}{c}\langle M^{p_n}_{{\delta_1},j},\pi^{p_n}_{{\delta_1},j}(p_n),\pi^{p_n}_{{\delta_1},j}({\delta_1}),\pi^{p_n}_{{\delta_1},j}(j),<^{p_n}_{{\delta_1},j}, G(N^{p_n}_{{\delta_1},j}\cap\omega_2)\rangle=\\
\langle M^{p_n}_{{\delta_2},j},\pi^{p_n}_{{\delta_2},j}(p_n),\pi^{p_n}_{{\delta_2},j}({\delta_2}),\pi^{p_n}_{{\delta_2},j}(j),<^{p_n}_{{\delta_2},j}, G(N^{p_n}_{{\delta_2},j}\cap\omega_2)\rangle\end{array}.$$
So $g_n:=(\pi^{p_n}_{\delta_2,j})^{-1}\circ \pi^{p_n}_{\delta_1,j}:N^{p_n}_{\delta_1,j}\rightarrow N^{p_n}_{\delta_2,j}$ 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 $x,y\in N^{p_n}_{\delta_1,j}$.
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 $G(N^{p_n}_{{\delta_1},j}\cap\omega_2)=G(N^{p_n}_{{\delta_2},j}\cap\omega_2)$ implies that $g_n\restriction (N^{p_n}_{{\delta_1},j}\cap N^{p_n}_{{\delta_2},j}\cap\omega_2)$ is the identity function.
Let $g:=\bigcup_{n<\omega}g_n$. Then $g$ is an isomorphism from $N_1$ to $N_2$, and $g\restriction( N_1\cap N_2\cap\omega_2)$ is the identity function. $\blacksquare$
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, $\gamma_1\neq\gamma_2$ 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 $\gamma_1\in N^{p_n}_{\delta_1,j}$, there would have existed a countable set $A\subseteq \gamma_1\cap N^{p_n}_{\delta_1,j}$ which is cofinal in $\gamma_1$. Recalling that $g\restriction N_1\cap\gamma_1$ is the identity function, we would have then get that $A\subseteq N^{p_n}_{\delta_2,j}$, and then $\gamma_1$ would have been an element of $N^{p_{n+1}}_{\delta_2,j}$ (note that $\gamma_2\in N^{p_n}_{\delta_2,j}$ and $\sup(N^{p_n}_{\delta_2,j}\cap\gamma_2)=\gamma_1$), contradicting the fact that $\gamma_1\not\in N_2$.
So $\gamma_1$ and $\gamma_2$ are distinct elements of $S$. Since $g(\delta_1)=\delta_2\neq\delta_1$, we get that $\delta_1\not\in N_2$ and $\delta_2\not\in N_1$. 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 $g\restriction N_1\cap\gamma_1$ 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, $\gamma_2\in T^{\gamma_1}_l$. 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 $c_{\delta_1}\restriction l+1=c_{\delta_2}\restriction l+1$, and hence $l\le j$. On the other hand, $j\in N^{p_0}_{\delta_1,j}$, and hence $j<N_1\cap\omega_2=l$. This is a contradiction. $\square$
