w23-logic-2/inputs/lecture_17.tex
Josia Pietsch 1b02d7e7f3
Some checks failed
Build latex and deploy / checkout (push) Failing after 13m44s
fixed numbers of lectures and failed merge
2024-01-10 22:30:24 +01:00

220 lines
7.9 KiB
TeX

\lecture{17}{2023-12-14}{Silver's Theorem}
We now want to prove \yaref{thm:silver}.
More generally, if $\kappa$ is a singular cardinal of uncountable cofinality
such that $2^{\lambda} = \lambda^+$ for all $\lambda < \kappa$,
then $2^{\kappa} = \kappa^+$.
\begin{remark}
The hypothesis of \yaref{thm:silver}
is consistent with $\ZFC$.
\end{remark}
We will only proof \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$.
The general proof differs only in notation.
\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:silver}
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 $\alpha \in T$, $h_Y(\alpha) = \beta$.
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
\[
\{Y \subseteq \aleph_{\omega_1} : \exists i < \aleph_{\omega_1 + 1} .~Y \le X_i\}
= \bigcup_{i < \aleph_{\omega_1 + 1}} \{Y \subseteq \aleph_{\omega_1} : Y \le X_i\}
\]
has size $\le \aleph_{\omega_1 + 1}$
(in fact the size is exactly $\aleph_{\omega_1 + 1}$).
But
\[
\{Y \subseteq \aleph_{\omega_1} : \exists i < \aleph_{\omega_1 + 1} .~Y \le X_i\} = \cP(\aleph_{\omega_1 + 1})
\]
because if $X \subseteq \aleph_{\omega_1 + 1}$
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}.
\end{refproof}