Josia Pietsch
ab14be172f
Some checks failed
Build latex and deploy / checkout (push) Failing after 14m16s
277 lines
12 KiB
TeX
277 lines
12 KiB
TeX
\lecture{17}{2023-12-14}{Silver's Theorem}
|
|
|
|
We now want to prove \yaref{thm:silver}.
|
|
|
|
\gist{%
|
|
\begin{remark}
|
|
The hypothesis of \yaref{thm:silver}
|
|
is consistent with $\ZFC$.
|
|
\end{remark}
|
|
}{}
|
|
|
|
We will only prove
|
|
\gist{%
|
|
\yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$
|
|
(see \yaref{thm:silver1}).
|
|
The general proof differs only in notation.
|
|
}{\yaref{thm:silver1}.}
|
|
\gist{%
|
|
\begin{remark}
|
|
It is important that the cofinality is uncountable.
|
|
For example it is consistent
|
|
with $\ZFC$ that
|
|
$2^{\aleph_n} = \aleph_{n+1}$ for all $n < \omega$
|
|
but at the same time $2^{\aleph_{\omega}} = \aleph_{\omega + 2}$.
|
|
\end{remark}
|
|
}{}
|
|
|
|
\begin{refproof}{thm:silver1}
|
|
\gist{%
|
|
We need to count the number of $X \subseteq \aleph_{\omega_1}.$
|
|
Let us fix $\langle f_\lambda : \lambda < \kappa \text{ an infinite cardinal} \rangle$
|
|
such that $f_{\lambda}\colon \cP(\lambda) \to \lambda^+$
|
|
is bijective for each $\lambda < \kappa$.
|
|
|
|
For $X \subseteq \aleph_{\omega_1}$
|
|
define
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
f_X\colon \omega_1 &\longrightarrow & \aleph_{\omega_1} \\
|
|
\alpha &\longmapsto & f_{\aleph_\alpha}(X \cap \aleph_\alpha).
|
|
\end{IEEEeqnarray*}
|
|
|
|
\begin{claim}
|
|
For $X,Y \subseteq \aleph_{\omega_1}$
|
|
it is $X \neq Y \iff f_X \neq f_Y$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
$X \neq Y$ holds iff $X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha$
|
|
for some $\alpha < \omega_1$.
|
|
But then $f_X(\alpha) \neq f_Y(\alpha)$.
|
|
\end{subproof}
|
|
|
|
For $X, Y \subseteq \aleph_{\omega_1}$
|
|
write $X \le Y$ iff
|
|
\[
|
|
\{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}
|
|
\]
|
|
is stationary.
|
|
|
|
\begin{claim}
|
|
For all $X,Y \subseteq \aleph_{\omega_1}$,
|
|
$X \le Y$ or $Y \le X$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
Suppose that $X \nleq Y$ and $Y \nleq X$.
|
|
Then there are clubs $C,D \subseteq \omega_1$
|
|
such that
|
|
\[
|
|
C \cap \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\} = \emptyset
|
|
\]
|
|
and
|
|
\[
|
|
D \cap \{\alpha < \omega_1 : f_Y(\alpha) \le f_X(\alpha)\} = \emptyset.
|
|
\]
|
|
Note that $C \cap D$ is a club.
|
|
Take some $\alpha \in C \cap D$.
|
|
But then $f_X(\alpha) \le f_Y(\alpha)$
|
|
or $f_{Y}(\alpha) \le f_X(\alpha)$ $\lightning$
|
|
\end{subproof}
|
|
|
|
\begin{claim}
|
|
\label{thm:silver:p:c3}.
|
|
Let $X \subseteq \aleph_{\omega_1}$.
|
|
Then
|
|
\[
|
|
|\{Y \subseteq \aleph_{\omega_1} : Y \le X\}| \le \aleph_{\omega_1}.
|
|
\]
|
|
\end{claim}
|
|
\begin{subproof}
|
|
Write $A \coloneqq \{Y \subseteq X_{\omega_1} : Y \le X\}$.
|
|
Suppose $|A| \ge \aleph_{\omega_1 + 1}$.
|
|
For each $Y \in A$
|
|
we have that
|
|
\[
|
|
S_Y \coloneqq \{\alpha : f_Y(\alpha) \le f_X(\alpha)\}
|
|
\]
|
|
is a stationary subset of $\omega_1$.
|
|
Since by assumption $2^{\aleph_1} = \aleph_2$,
|
|
there are at most $\aleph_2$ such $S_Y$.
|
|
|
|
Suppose that for each $S \subseteq \omega_1$,
|
|
\[
|
|
|\{Y \in A : S_Y = S\}| < \aleph_{\omega_1 + 1}.
|
|
\]
|
|
Then $A$ is the union of $\le \aleph_2$ many
|
|
sets of size $< \aleph_{\omega_1 + 1}$.
|
|
Thus this is a contradiction since $\aleph_{\omega_1 + 1}$
|
|
is regular.
|
|
|
|
So there exists a stationary $S \subseteq \omega_1$
|
|
such that
|
|
\[
|
|
A_1 = \{Y \subseteq \aleph_{\omega_1} : S_Y = S\}
|
|
\]
|
|
has cardinality $\aleph_{\omega_1 + 1}$.
|
|
We have
|
|
\[f_Y(\alpha) \le f_X(\alpha) = f_{\aleph_\alpha}(X \cap \aleph_\alpha)< \aleph_{\alpha + 1}\]
|
|
for all $Y \in A_1, \alpha \in S$.
|
|
|
|
Let $\langle g_{\alpha} : \alpha \in S \rangle$
|
|
be such that $g_\alpha\colon \aleph_{\alpha} \twoheadrightarrow f_X(\alpha) + 1$
|
|
is a surjection for all $\alpha \in S$.
|
|
|
|
Then for each $Y \in A_1$ define
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\overline{f}_Y\colon S &\longrightarrow & \aleph_{\omega_1} \\
|
|
\alpha &\longmapsto & \min \{\xi : g_\alpha(\xi) = f_Y(\alpha)\}.
|
|
\end{IEEEeqnarray*}
|
|
|
|
Let $D$ be the set of all limit ordinals $< \omega_1$.
|
|
Then $S \cap D$ is a stationary set:
|
|
If $C$ is a club, then $C \cap D$ is a club,
|
|
hence $(S \cap D) \cap C = S \cap (D \cap C) \neq \emptyset$.
|
|
|
|
Now to each $Y \in A$ we may associate
|
|
a regressive function
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
h_Y \colon S \cap D &\longrightarrow & \omega_1 \\
|
|
\alpha &\longmapsto & \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_{\beta}\}.
|
|
\end{IEEEeqnarray*}
|
|
|
|
$h_Y$ is regressive, so by \yaref{thm:fodor}
|
|
there is a stationary $T_Y \subseteq S \cap D$ on which $h_Y$ is constant.
|
|
|
|
By an argument as before,
|
|
there is a stationary $T \subseteq S \cap D$ such that
|
|
\[
|
|
|A_2| = \aleph_{\omega_1 +1},
|
|
\]
|
|
where $A_2 \coloneqq \{Y \in A_1 : T_Y = T\}$.
|
|
|
|
Let $\beta < \omega_1$ be such that for all $Y \in A_2$
|
|
and for all $\alpha \in T$, $h_Y(\alpha) = \beta$.
|
|
% TODO WHY DOES THIS WORK?
|
|
Then $\overline{f}_Y(\alpha) < \aleph_\beta$
|
|
for all $Y \in A_2$ and $\alpha \in T$.
|
|
|
|
There are at most $\aleph_\beta^{\aleph_1}$ many functions
|
|
$T \to \aleph_\beta$,
|
|
but
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\aleph_\beta^{\aleph_1} &\le & 2^{\aleph_\beta \cdot \aleph_1}\\
|
|
&=& \aleph_{\beta+1} \cdot \aleph_2\\
|
|
&<& \aleph_{\omega_1}.
|
|
\end{IEEEeqnarray*}
|
|
|
|
Suppose that for each function
|
|
$\tilde{f}\colon T \to \aleph_\beta$
|
|
there are $< \aleph_{\omega_1 + 1}$ many $Y \in A_2$
|
|
with $\overline{f}_Y \cap T = \tilde{f}$.
|
|
|
|
Then $A_2$ is the union of $<\aleph_{\omega_1}$
|
|
many sets each of size $< \aleph_{\omega_1 + 1}$ $\lightning$.
|
|
Hence for some $\tilde{f}\colon T \to \aleph_\beta$,
|
|
\[
|
|
|A_3| = \aleph_{\omega_1 + 1},
|
|
\]
|
|
where $A_3 = \{Y \in A_2 : \overline{f}_Y\defon{T} = \tilde{f}\}$.
|
|
|
|
Let $Y, Y' \in A_3$ and $\alpha \in T$.
|
|
Then
|
|
\[
|
|
\overline{f}_Y(\alpha) = \overline{f}_{Y'}(\alpha),
|
|
\]
|
|
hence
|
|
\[
|
|
f_{\aleph_\alpha}(Y \cap \aleph_\alpha) = f_Y(\alpha) = f_{Y'}(\alpha) = f_{\aleph_\alpha}(Y' \cap \aleph_\alpha),
|
|
\]
|
|
i.e.~$Y \cap \aleph_\alpha = Y' \cap \aleph_\alpha$.
|
|
Since $T$ is cofinal in $\omega_1$,
|
|
it follows that $Y = Y'$.
|
|
So $|A_3| \le 1 \lightning$
|
|
\end{subproof}
|
|
|
|
Let us now define a sequence $\langle X_i : i < \aleph_{\omega_1 + 1} \rangle$
|
|
of subsets of $\aleph_{\omega_1 + 1}$ as follows:
|
|
|
|
Suppose $\langle X_j : j < i \rangle$
|
|
were already chosen.
|
|
Consider
|
|
\[
|
|
\{Y \subseteq \aleph_{\omega_1} : \exists j < i.~Y \le X_j\}
|
|
= \bigcup_{j < i} \{Y \subseteq \aleph_{\omega_1} : Y \le X_j\}.
|
|
\]
|
|
This set has cardinality $\le \aleph_{\omega_1}$
|
|
by \yaref{thm:silver:p:c3}.
|
|
Let $X_i \subseteq \aleph_{\omega_1}$
|
|
be such that $X_i \nleq X_j$ for all $j < i$.
|
|
|
|
|
|
The set
|
|
\[
|
|
P \coloneqq \{Y \subseteq \aleph_{\omega_1} : \exists i < \aleph_{\omega_1 + 1} .~Y \le X_i\}
|
|
= \bigcup_{i < \aleph_{\omega_1 + 1}} \{Y \subseteq \aleph_{\omega_1} : Y \le X_i\}
|
|
\]
|
|
has size $\le \aleph_{\omega_1 + 1}$
|
|
(in fact the size is exactly $\aleph_{\omega_1 + 1}$).
|
|
|
|
On the other hand
|
|
$P = \cP(\aleph_{\omega_1})$
|
|
because if $X \subseteq \aleph_{\omega_1}$
|
|
is such that $X \nleq X_i$ for all $i < \aleph_{\omega_1 + 1}$,
|
|
then $X_i \le X$ for all $i < \aleph_{\omega_1 + 1}$,
|
|
so such a set $X$ does not exist by \yaref{thm:silver:p:c3}.
|
|
}{Need to count $X \subseteq \aleph_{ \omega_{1}}$.
|
|
\begin{itemize}
|
|
\item Fix bijections $f_\lambda\colon 2^{\lambda} \to \lambda^+$
|
|
for all infinite cardinals $\lambda < \kappa$.
|
|
\item For $X \subseteq \aleph_{ \omega_1}$ define $f_X\colon \omega_1 \to \aleph_{ \omega_1},
|
|
\alpha \mapsto f_{\aleph_\alpha}(X \cap \aleph_\alpha)$.
|
|
\item (1) $X \neq Y \iff f_X \neq f_Y$:
|
|
\begin{itemize}
|
|
\item $X \neq Y \iff \exists \alpha.~X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha \iff \exists \alpha.~f_X(\alpha) \neq f_Y(\alpha)$.
|
|
\end{itemize}
|
|
\item $X \le Y :\iff \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}$ stationary.
|
|
\item (2) $X\le Y \lor Y \le X$:
|
|
Suppose $X \not\le Y, Y \not\le X$. Choose witnessing clubs $C, D$.
|
|
$C \cap D$ is club, so $C\cap D \ni \alpha \implies f_X(\alpha) \substack{\nleq\\\ngeq} f_Y(\alpha) \lightning$
|
|
\item (3) $X \subseteq \aleph_{ \omega_1}$, then $|\underbrace{\{Y \subseteq \aleph_{ \omega_1} : Y \le X\}}_{A}| \le \aleph_{ \omega_1}$
|
|
\begin{itemize}
|
|
\item Suppose $|A| \ge \aleph_{ \omega_1 + 1}$.
|
|
\item $S_Y \coloneqq \{\alpha : f_Y(\alpha) \le f_X(\alpha)\}$ stationary for all $Y \in A$.
|
|
$2^{\aleph_1} = \aleph_2 \implies$ at most $\aleph_2$ such $S_Y$.
|
|
\item If $\forall S \in \cP( \omega_1).~ |\underbrace{\{Y \in A : S_Y = S\}}_{A_S}| < \aleph_{ \omega_1 + 1}$,
|
|
then $|A| \le \aleph_2 \cdot <\aleph_{ \omega_1 + 1}$ $\lightning$ $\aleph_{ \omega_1 + 1}$ regular.
|
|
\item So $\exists S \in \omega_1.~|A_S| = \aleph_{ \omega_1 + 1}$.
|
|
\item Fix surjection $\langle g_\alpha : \aleph_\alpha \twoheadrightarrow f_X(\alpha) + 1 : \alpha \in S \rangle$.
|
|
($f_Y(\alpha) \le f_X(\alpha) < \aleph_{\alpha+1}$)
|
|
\item $\forall Y \in A_S$ define $\overline{f}_Y \colon S \to \aleph_{ \omega_1}, \alpha \mapsto \min \{\xi : g_\alpha(\xi) = f_Y(\alpha)\}$.
|
|
\item $S^\ast$ ($S \cap$ limit ordinals) is stationary.
|
|
\item $\forall Y \in A$ define $h_Y\colon S^\ast \to \omega_1, \alpha \mapsto \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_\beta\}$.
|
|
\item Apply \yaref{thm:fodor} to $h_Y, S^\ast$ to get $T_Y \subseteq S^\ast$ stationary with $h_Y\defon{T_Y}$ constant.
|
|
\item $\exists T.~|\underbrace{\{Y \in A_S : T_Y = T\}}_{A_{S,T}}| = \aleph_{ \omega_1 + 1}$.
|
|
\item Let $\{\beta\} = h_Y''T$, i.e.~$\overline{f}_Y(\alpha) < \aleph_{\beta}$ for $Y\in A_{S,T}, \alpha \in T$.
|
|
\item $\leftindex^T \aleph_\beta \le 2^{\aleph_\beta \cdot \aleph_1} = \aleph_{\beta+1} \aleph_2 < \aleph_{ \omega_1}$.
|
|
\item $\exists \tilde{f}\colon T \to \aleph_\beta .~ |\underbrace{\{Y \in A_{S,T} : \overline{f}_Y\defon{T} = \tilde{f}\} }_{A_{S,T,\tilde{f}}}| = \aleph_{ \omega_1 + 1}$.
|
|
\item $Y,Y' \in A_{S,T,\tilde{f}} \implies \forall \alpha \in T.~f_Y(\alpha) = f_{Y'}\left( \alpha \right) \overset{T \text{ unbounded}}{\implies} Y = Y'$
|
|
Thus $|A_{S,T, \tilde{f}} | \le 1 \lightning$.
|
|
\end{itemize}
|
|
|
|
\item Define sequence $\langle X_i : i < \aleph_{ \omega_1 + 1} \rangle$
|
|
of subsets of $\aleph_{\omega_1}$:
|
|
\begin{itemize}
|
|
\item Consider $\{Y \subseteq \aleph_{ \omega_1} : \exists j < i.~Y \le X_j\}$
|
|
(cardinality $ \le \aleph_{ \omega_1}$),
|
|
Take $X_i \subseteq \aleph_{ \omega_1}$
|
|
such that $X_i \subseteq \aleph_{ \omega_1}$
|
|
such that $X_i \not\le X_j$ for all $j < i$.
|
|
\item $P \coloneqq \{Y \subseteq \aleph_{ \omega_1} : \exists i < \aleph_{ \omega_1 + 1}.~Y \le X_i\}$
|
|
\end{itemize}
|
|
\item $|P| \overset{\text{in fact } =}{\le} \aleph_{ \omega_1 + 1} \aleph_{ \omega_1} = \aleph_{ \omega_1 + 1}$ by (3).
|
|
\item $X \in \cP(\aleph_{ \omega_1}) \setminus P \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
|
|
$\lightning$ (3).
|
|
Thus $P = \cP(\aleph_{ \omega_1})$.
|
|
\end{itemize}
|
|
}
|
|
|
|
\end{refproof}
|