fixed numbers of lectures and failed merge
Some checks failed
Build latex and deploy / checkout (push) Failing after 13m44s

This commit is contained in:
Josia Pietsch 2024-01-10 22:30:24 +01:00
parent 8ce92917ed
commit 1b02d7e7f3
Signed by: josia
GPG key ID: E70B571D66986A2D
3 changed files with 221 additions and 227 deletions

View file

@ -1,4 +1,3 @@
<<<<<<< HEAD
\lecture{16}{2023-12-11}{} \lecture{16}{2023-12-11}{}
Recall \yaref{thm:fodor}. Recall \yaref{thm:fodor}.
@ -238,227 +237,3 @@ then it is true at $\kappa$.
The proof of \yalabel{thm:silver} is quite elementary, The proof of \yalabel{thm:silver} is quite elementary,
so we will do it now, but the statement can only be fully appreciated later. so we will do it now, but the statement can only be fully appreciated later.
||||||| ede97ee
=======
\lecture{16}{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}
>>>>>>> 5ee970194987e24c54bfc9c787cb219e15e71520

220
inputs/lecture_17.tex Normal file
View file

@ -0,0 +1,220 @@
\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}

View file

@ -1,5 +1,4 @@
\lecture{17}{2023-12-18}{Large cardinals} \lecture{18}{2023-12-18}{Large cardinals}
\begin{definition} \begin{definition}
\begin{itemize} \begin{itemize}