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2
inputs/lecture_07.tex
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@ -9,7 +9,7 @@
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item $0$ is an ordinal, and if $\alpha$ is \item $0$ is an ordinal, and if $\alpha$ is
an ordinal, so is $\alpha + 1$. an ordinal, so is $\alpha + 1$.
\item If $\alpha$ is an ordinal and $x \in \alpha$, \item If $\alpha$ is an ordinal and $x \in a$,
then $x$ is an ordinal. then $x$ is an ordinal.
\item If $\alpha, \beta$ are ordinals \item If $\alpha, \beta$ are ordinals
and $\alpha \subseteq \beta$, and $\alpha \subseteq \beta$,

2
inputs/lecture_09.tex
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@ -54,7 +54,7 @@ An alternative way of formulating this is
Let $D$ be a class of triples Let $D$ be a class of triples
such that for all $u,x$ there is exactly such that for all $u,x$ there is exactly
one $y$ with $(u,x,y) \in D$ one $y$ with $(u,x,y) \in D$
(basically $(u,x) \mapsto y$ is a function). (basicalls $(u,x) \mapsto y$ is a function).
Then there is a unique function $f$ on $A$ Then there is a unique function $f$ on $A$
such that for all $x \in A$, such that for all $x \in A$,

2
inputs/lecture_10.tex
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@ -158,7 +158,7 @@ This $F$ is called the \vocab{rank function} for $(A, R)$.
\] \]
and and
\[ \[
\rank(R) \coloneqq \ran(F). \rk_R \rank(R) \coloneqq \ran(F).
\] \]
In the special case that $R$ is a linear order on $A$, In the special case that $R$ is a linear order on $A$,

225
inputs/lecture_16.tex
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@ -1,3 +1,4 @@
<<<<<<< HEAD
\lecture{16}{2023-12-11}{} \lecture{16}{2023-12-11}{}
Recall \yaref{thm:fodor}. Recall \yaref{thm:fodor}.
@ -237,3 +238,227 @@ 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

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inputs/lecture_17.tex
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@ -1,220 +1,240 @@
\lecture{17}{2023-12-14}{Silver's Theorem} \lecture{17}{2023-12-18}{Large cardinals}
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{definition}
\begin{itemize}
\item A cardinal $\kappa$ is called \vocab{weakly inaccessible}
iff $\kappa$ is uncountable,\footnote{dropping this we would get that $\aleph_0$ is inaccessible}
regular and $\forall \lambda < \kappa.~\lambda^+ < \kappa$.
\item A cardinal $\kappa$ is \vocab[inaccessible!strongly]{(strongly) inaccessible}
iff $\kappa$ is uncountable, regular
and $\forall \lambda < \kappa.~2^{\lambda} < \kappa$.
\end{itemize}
\end{definition}
\begin{remark} \begin{remark}
The hypothesis of \yaref{thm:silver} Since $2^{\lambda} \ge \lambda^+$,
is consistent with $\ZFC$. strongly inaccessible cardinals are weakly inaccessible.
If $\GCH$ holds, the notions coincide.
\end{remark} \end{remark}
We will only proof \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$. \begin{theorem}
The general proof differs only in notation. If $\kappa$ is inaccessible,
then $V_\kappa \models \ZFC$.\footnote{More formally $(V_{\kappa}, \in ) \models\ZFC$.}
\end{theorem}
\begin{proof}
Since $\kappa$ is regular, \AxRep works.
Since $2^{\lambda} < \kappa$,
\AxPow works.
The other axioms are trivial.
\todo{Exercise}
\end{proof}
\begin{corollary}
$\ZFC$ does not prove the existence of inaccessible
cardinals, unless $\ZFC$ is inconsistent.
\end{corollary}
\begin{proof}
If $\ZFC$ is consistent,
it can not prove that it is consistent.
In particular, it can not prove the existence of a model of $\ZFC$.
\end{proof}
\begin{remark} \begin{definition}[Ulam]
It is important that the cofinality is uncountable. A cardinal $\kappa > \aleph_0$ is \vocab{measurable}
For example it is consistent iff there is an ultrafilter $U$ on $\kappa$,
with $\ZFC$ that such that $U$ is not principal\footnote{%
$2^{\aleph_n} = \aleph_{n+1}$ for all $n < \omega$ i.e.~$\{\xi\} \not\in U$ for all $\xi < \kappa$%
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 and
\[ if $\theta < \kappa$
D \cap \{\alpha < \omega_1 : f_Y(\alpha) \le f_X(\alpha)\} = \emptyset. and $\{X_i : i < \theta\} \subseteq U$,
\] then $\bigcap_{i < \theta} X_i \in U$
Note that $C \cap D$ is a club. \end{definition}
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{goal}
We want to prove
that if $\kappa$ is measurable,
then $\kappa$ is inaccessible
and there are $\kappa$ many
inaccessible cardinals below $\kappa$
(i.e.~$\kappa$ is the $\kappa$\textsuperscript{th} inaccessible).
\end{goal}
\begin{theorem}
The following are equivalent:
\begin{enumerate}
\item $\kappa$ is a measurable cardinal.
\item There is an elementary embedding%
\footnote{Recall: $j\colon V \to M$
is an \vocab{elementary embedding} iff $j''V = \{j(x) : x \in V\} \prec M$,
i.e.~for all formulae $\phi$ and $x_1,\ldots,x_k \in V$,
$V \models\phi(x_1,\ldots,x_u) \iff M \models\phi(j(x_1),\ldots,j(x_u))$.%
}
$j\colon V \to M$ with $M$
transitive
such that $j\defon{\kappa} = \id$,
$j(\kappa) \neq \kappa$.
\end{enumerate}
\end{theorem}
\begin{proof}
2. $\implies$ 1.:
Fox $j\colon V \to M$.
Let $U = \{X \subseteq \kappa : \kappa \in j(X)\}$.
We need to show that $U$ is an ultrafilter:
\begin{itemize}
\item Let $X,Y \in U$.
Then $\kappa \in j(X) \cap j(Y)$.
We have $M \models j(X \cap Y) = j(X) \cap j(Y)$,
and thus $j(X \cap Y) = j(X) \cap j(Y)$.
It follows that $X \cap Y \in U$.
\item Let $X\in U$
and $X \subseteq Y \subseteq \kappa$.
Then $ \kappa \in j(X) \subseteq j(Y)$
by the same argument,
so $Y \in U$.
\item We have $j(\emptyset) = \emptyset$
(again $M \models j(\emptyset)$ is empty),
hence $\emptyset\not\in U$.
\item $\kappa \in U$ follows from $\kappa \in j(\kappa)$.
This is shown as follows:
\begin{claim} \begin{claim}
\label{thm:silver:p:c3}. For every ordinal $\alpha$,
Let $X \subseteq \aleph_{\omega_1}$. $j(\alpha)$ is an ordinal such that $j(\alpha) \ge \alpha$.
Then
\[
|\{Y \subseteq \aleph_{\omega_1} : Y \le X\}| \le \aleph_{\omega_1}.
\]
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
Write $A \coloneqq \{Y \subseteq X_{\omega_1} : Y \le X\}$. $\alpha \in \Ord$
Suppose $|A| \ge \aleph_{\omega_1 + 1}$. can be written as
For each $Y \in A$ \[\forall x \in \alpha .~\forall y \in x.~y \in \alpha
we have that \land \forall x \in \alpha .~ \forall y \in \alpha.~(x \in y \lor x = y \lor y \in x).
\[
S_Y \coloneqq \{\alpha : f_Y(\alpha) \le f_X(\alpha)\}
\] \]
is a stationary subset of $\omega_1$. So if $\alpha$ is an ordinal,
Since by assumption $2^{\aleph_1} = \aleph_2$, then $M \models \text{``$j(\alpha)$ is an ordinal''}$
there are at most $\aleph_2$ such $S_Y$. in the sense above.
Therefore $j(\alpha)$ really is an ordinal.
Suppose that for each $S \subseteq \omega_1$, If the claim fails,
\[ we can pick the least $\alpha$ such that $j(\alpha) < \alpha$.
|\{Y \in A : S_Y = S\}| < \aleph_{\omega_1 + 1}. Then $M \models j(j(\alpha)) < j(\alpha)$,
\] i.e. $j(j(\alpha)) < j(\alpha)$
Then $A$ is the union of $\le \aleph_2$ many contradicting the minimality of $\alpha$.
sets of size $< \aleph_{\omega_1 + 1}$. \end{subproof}
Thus this is a contradiction since $\aleph_{\omega_1 + 1}$
is regular.
So there exists a stationary $S \subseteq \omega_1$ Therefore as $j(\kappa) \neq \kappa$,
such that we have $j(\kappa) > \kappa$,
\[ i.e.~$\kappa \in j(\kappa)$.
A_1 = \{Y \subseteq \aleph_{\omega_1} : S_Y = S\} \item $U$ is an ultrafilter:
\] Let $X \subseteq \kappa$.
has cardinality $\aleph_{\omega_1 + 1}$. Then $\kappa \in j(\kappa) = j(X \cup (\kappa \setminus X)) = j(X) \cup j(\kappa \setminus X)$.
We have So $X \in U$ or $\kappa \setminus X \in U$.
\[f_Y(\alpha) \le f_X(\alpha) = f_{\aleph_\alpha}(X \cap \aleph_\alpha)< \aleph_{\alpha + 1}\]
for all $Y \in A_1, \alpha \in S$.
Let $\langle g_{\alpha} : \alpha \in S \rangle$ Let $\theta < \kappa$
be such that $g_\alpha\colon \aleph_{\alpha} \twoheadrightarrow f_X(\alpha) + 1$ and $\{X_i : i < \theta\} \subseteq U$.
is a surjection for all $\alpha \in S$. Then $\kappa \in j(X_i)$ for all $i < \theta$,
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 hence
\[ \[
f_{\aleph_\alpha}(Y \cap \aleph_\alpha) = f_Y(\alpha) = f_{Y'}(\alpha) = f_{\aleph_\alpha}(Y' \cap \aleph_\alpha), \kappa \in \bigcap_{i < \theta} j(X_i)
= j\left( \bigcap_{i < \theta} X_i \right) \in U.
\] \]
i.e.~$Y \cap \aleph_\alpha = Y' \cap \aleph_\alpha$. This holds since $j(\theta) = \theta$ (as $\theta < \kappa$),
Since $T$ is cofinal in $\omega_1$, so $j(\langle X_i : i < \theta \rangle) = \langle j(X_i) : i < \theta \rangle$.
it follows that $Y = Y'$.
So $|A_3| \le 1 \lightning$
\end{subproof}
Let us now define a sequence $\langle X_i : i < \aleph_{\omega_1 + 1} \rangle$ Also if $\xi < \kappa$,
of subsets of $\aleph_{\omega_1 + 1}$ as follows: then $j(\{\xi\}) = \{\xi\}$
so $\kappa \not\in j(\{\xi\})$
and $\{\xi\} \not\in U$.
\end{itemize}
Suppose $\langle X_j : j < i \rangle$
were already chosen. 1. $\implies$ 2.
Consider Fix $U$.
Let ${}^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
For $f,g \in {}^{\kappa}V$ define
$f \sim g :\iff \{\xi < \kappa : f(\xi) = g(\xi)\} \in U$.
This is an equivalence relation since $U$ is a filter.
Write $[f] = \{g \in {}^\kappa V : g \sim ~ f \land g \in V_\alpha \text{ for the least $\alpha$ such that there is some $h \in V_\alpha$ with $h \sim f$}\}$.%
\footnote{This is know as \vocab{Scott's Trick}.
Note that by defining equivalence classes
in the usual way (i.e.~without this trick),
one ends up with proper classes:
For $f\colon \kappa \to V$,
we can for example change $f(0)$ to be an arbitrary $V_{\alpha}$
and get another element of $[f]$.
}
For any two such equivalence classes $[f], [g]$
define
\[[f] \tilde{\in} [g] :\iff \{\xi < \kappa : f(\xi) \in g(\xi)\} \in U.\]
This is independent of the choice of the representatives,
so it is well-defined.
Now write $\cF = \{[f] : f \in {}^{\kappa}V\}$
and look at $(\cF, \tilde{\in})$.
The key to the construction is \yaref{thm:los} (see below).
Given \yaref{thm:los},
we may define an elementary embedding
$\overline{j}\colon (V, \in ) \to (\cF, \tilde{\in })$
as follows:
Let $\overline{j}(x) = [c_x]$,
where $c_x \colon \kappa \to \{x\}$ is the constant function
with value $x$.
Then
\begin{IEEEeqnarray*}{rCl}
(V, \in ) \models \phi(x_1,\ldots,x_k)&\iff&
\{\xi < \kappa: (V, \in ) \models \phi(c_{x_1}(\alpha), \ldots, c_{x_k}(\alpha))\} \in U\\
&\overset{\yaref{thm:los}}{\iff}&
(\cF, \tilde{\in }) \models \phi(\overline{j}(x_1), \ldots, \overline{j}(x_k)).
\end{IEEEeqnarray*}
Let us show that $(\cF, \tilde{\in })$
is well-founded.
Otherwise there is $\langle f_n : n < \omega \rangle$
such that $f_n \in {}^\kappa V$
and $[f_{n+1}] \tilde{\in } [f_n]$ for all $n < \omega$.
Then $X_n \coloneqq \{\xi < \kappa : f_{n+1}(\xi) \in f_n(\xi)\} \in U$,
so $\bigcap X_n \in U$.
Let $\xi_0 \in \bigcap X_n$.
Then $f_0(\xi_0) \ni f_1(\xi_0) \ni f_2(\xi_0) \ni \ldots$ $\lightning$.
Note that $\tilde{\in }$ is set-like,
therefore by the \yaref{lem:mostowski}
there is some transitive $M$ with $(\cF, \tilde{\in }) \overset{\sigma}{\cong} (M, \in )$.
We can now define an elementary embedding $j\colon V \to M$
by $j \coloneqq \sigma \circ \overline{j}$.
It remains to show that $\alpha < \kappa \implies j(\alpha) = \alpha$.
This can be done by induction:
Fix $\alpha$. We already know $j(\alpha) \ge \alpha$.
Suppose $\beta \in j(\alpha)$.
Then $\beta = \sigma([f])$ for some $f$
and $\sigma([f]) \in \sigma([c_{\alpha}])$,
i.e.~$[f] \tilde{\in} [c_\alpha]$.
Thus $\{\xi < \kappa : f(\xi) \in \underbrace{c_{\alpha}(\xi)}_{\alpha}\} \in U$.
Hence there is some $\delta < \alpha$
such that
\[
X_\delta \coloneqq \{\xi < \kappa : f(\xi) = \delta\} \in U,
\]
as otherwise $\forall \delta < \alpha. ~ \kappa \setminus X_\delta \in U$,
i.e.~$\emptyset = (\bigcap_{\delta < \alpha} \kappa \setminus X_{\delta}) \cap X \in U \lightning$.
We get $[f] = [c_{\delta}]$,
so $\beta = \sigma([f]) = \delta([c_{\delta}]) = j(\delta) = \delta$,
where for the last equality we have applied the induction hypothesis.
So $j(\alpha) \le \alpha$.
% It is also easy to show $j(\kappa) > \kappa$.
\end{proof}
\begin{theorem}[\L o\'s]
\yalabel{\L o\'s's Theorem}{\L o\'s}{thm:los}
For all formulae $\phi$ and for all $f_1, \ldots, f_k \in {}^{\kappa} V$,
\[ \[
\{Y \subseteq \aleph_{\omega_1} : \exists j < i.~Y \le X_j\} (\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
= \bigcup_{j < i} \{Y \subseteq \aleph_{\omega_1} : Y \le X_j\}. \iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.
\] \]
This set has cardinality $\le \aleph_{\omega_1}$ \end{theorem}
by \yaref{thm:silver:p:c3}. \begin{proof}
Let $X_i \subseteq \aleph_{\omega_1}$ Induction on the complexity of $\phi$.
be such that $X_i \nleq X_j$ for all $j < i$. \end{proof}
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}

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\lecture{18}{2023-12-18}{Large cardinals}
\begin{definition}
\begin{itemize}
\item A cardinal $\kappa$ is called \vocab{weakly inaccessible}
iff $\kappa$ is uncountable,\footnote{dropping this we would get that $\aleph_0$ is inaccessible}
regular and $\forall \lambda < \kappa.~\lambda^+ < \kappa$.
\item A cardinal $\kappa$ is \vocab[inaccessible!strongly]{(strongly) inaccessible}
iff $\kappa$ is uncountable, regular
and $\forall \lambda < \kappa.~2^{\lambda} < \kappa$.
\end{itemize}
\end{definition}
\begin{remark}
Since $2^{\lambda} \ge \lambda^+$,
strongly inaccessible cardinals are weakly inaccessible.
If $\GCH$ holds, the notions coincide.
\end{remark}
\begin{theorem}
If $\kappa$ is inaccessible,
then $V_\kappa \models \ZFC$.\footnote{More formally $(V_{\kappa}, \in ) \models\ZFC$.}
\end{theorem}
\begin{proof}
Since $\kappa$ is regular, \AxRep works.
Since $2^{\lambda} < \kappa$,
\AxPow works.
The other axioms are trivial.
\todo{Exercise}
\end{proof}
\begin{corollary}
$\ZFC$ does not prove the existence of inaccessible
cardinals, unless $\ZFC$ is inconsistent.
\end{corollary}
\begin{proof}
If $\ZFC$ is consistent,
it can not prove that it is consistent.
In particular, it can not prove the existence of a model of $\ZFC$.
\end{proof}
\begin{definition}[Ulam]
A cardinal $\kappa > \aleph_0$ is \vocab{measurable}
iff there is an ultrafilter $U$ on $\kappa$,
such that $U$ is not principal\footnote{%
i.e.~$\{\xi\} \not\in U$ for all $\xi < \kappa$%
}
and
if $\theta < \kappa$
and $\{X_i : i < \theta\} \subseteq U$,
then $\bigcap_{i < \theta} X_i \in U$
\end{definition}
\begin{goal}
We want to prove
that if $\kappa$ is measurable,
then $\kappa$ is inaccessible
and there are $\kappa$ many
inaccessible cardinals below $\kappa$
(i.e.~$\kappa$ is the $\kappa$\textsuperscript{th} inaccessible).
\end{goal}
\begin{theorem}
The following are equivalent:
\begin{enumerate}
\item $\kappa$ is a measurable cardinal.
\item There is an elementary embedding%
\footnote{Recall: $j\colon V \to M$
is an \vocab{elementary embedding} iff $j''V = \{j(x) : x \in V\} \prec M$,
i.e.~for all formulae $\phi$ and $x_1,\ldots,x_k \in V$,
$V \models\phi(x_1,\ldots,x_u) \iff M \models\phi(j(x_1),\ldots,j(x_u))$.%
}
$j\colon V \to M$ with $M$
transitive
such that $j\defon{\kappa} = \id$,
$j(\kappa) \neq \kappa$.
\end{enumerate}
\end{theorem}
\begin{proof}
2. $\implies$ 1.:
Fox $j\colon V \to M$.
Let $U = \{X \subseteq \kappa : \kappa \in j(X)\}$.
We need to show that $U$ is an ultrafilter:
\begin{itemize}
\item Let $X,Y \in U$.
Then $\kappa \in j(X) \cap j(Y)$.
We have $M \models j(X \cap Y) = j(X) \cap j(Y)$,
and thus $j(X \cap Y) = j(X) \cap j(Y)$.
It follows that $X \cap Y \in U$.
\item Let $X\in U$
and $X \subseteq Y \subseteq \kappa$.
Then $ \kappa \in j(X) \subseteq j(Y)$
by the same argument,
so $Y \in U$.
\item We have $j(\emptyset) = \emptyset$
(again $M \models j(\emptyset)$ is empty),
hence $\emptyset\not\in U$.
\item $\kappa \in U$ follows from $\kappa \in j(\kappa)$.
This is shown as follows:
\begin{claim}
For every ordinal $\alpha$,
$j(\alpha)$ is an ordinal such that $j(\alpha) \ge \alpha$.
\end{claim}
\begin{subproof}
$\alpha \in \Ord$
can be written as
\[\forall x \in \alpha .~\forall y \in x.~y \in \alpha
\land \forall x \in \alpha .~ \forall y \in \alpha.~(x \in y \lor x = y \lor y \in x).
\]
So if $\alpha$ is an ordinal,
then $M \models \text{``$j(\alpha)$ is an ordinal''}$
in the sense above.
Therefore $j(\alpha)$ really is an ordinal.
If the claim fails,
we can pick the least $\alpha$ such that $j(\alpha) < \alpha$.
Then $M \models j(j(\alpha)) < j(\alpha)$,
i.e. $j(j(\alpha)) < j(\alpha)$
contradicting the minimality of $\alpha$.
\end{subproof}
Therefore as $j(\kappa) \neq \kappa$,
we have $j(\kappa) > \kappa$,
i.e.~$\kappa \in j(\kappa)$.
\item $U$ is an ultrafilter:
Let $X \subseteq \kappa$.
Then $\kappa \in j(\kappa) = j(X \cup (\kappa \setminus X)) = j(X) \cup j(\kappa \setminus X)$.
So $X \in U$ or $\kappa \setminus X \in U$.
Let $\theta < \kappa$
and $\{X_i : i < \theta\} \subseteq U$.
Then $\kappa \in j(X_i)$ for all $i < \theta$,
hence
\[
\kappa \in \bigcap_{i < \theta} j(X_i)
= j\left( \bigcap_{i < \theta} X_i \right) \in U.
\]
This holds since $j(\theta) = \theta$ (as $\theta < \kappa$),
so $j(\langle X_i : i < \theta \rangle) = \langle j(X_i) : i < \theta \rangle$.
Also if $\xi < \kappa$,
then $j(\{\xi\}) = \{\xi\}$
so $\kappa \not\in j(\{\xi\})$
and $\{\xi\} \not\in U$.
\end{itemize}
1. $\implies$ 2.
Fix $U$.
Let ${}^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
For $f,g \in {}^{\kappa}V$ define
$f \sim g :\iff \{\xi < \kappa : f(\xi) = g(\xi)\} \in U$.
This is an equivalence relation since $U$ is a filter.
Write $[f] = \{g \in {}^\kappa V : g \sim ~ f \land g \in V_\alpha \text{ for the least $\alpha$ such that there is some $h \in V_\alpha$ with $h \sim f$}\}$.%
\footnote{This is know as \vocab{Scott's Trick}.
Note that by defining equivalence classes
in the usual way (i.e.~without this trick),
one ends up with proper classes:
For $f\colon \kappa \to V$,
we can for example change $f(0)$ to be an arbitrary $V_{\alpha}$
and get another element of $[f]$.
}
For any two such equivalence classes $[f], [g]$
define
\[[f] \tilde{\in} [g] :\iff \{\xi < \kappa : f(\xi) \in g(\xi)\} \in U.\]
This is independent of the choice of the representatives,
so it is well-defined.
Now write $\cF = \{[f] : f \in {}^{\kappa}V\}$
and look at $(\cF, \tilde{\in})$.
The key to the construction is \yaref{thm:los} (see below).
Given \yaref{thm:los},
we may define an elementary embedding
$\overline{j}\colon (V, \in ) \to (\cF, \tilde{\in })$
as follows:
Let $\overline{j}(x) = [c_x]$,
where $c_x \colon \kappa \to \{x\}$ is the constant function
with value $x$.
Then
\begin{IEEEeqnarray*}{rCl}
(V, \in ) \models \phi(x_1,\ldots,x_k)&\iff&
\{\xi < \kappa: (V, \in ) \models \phi(c_{x_1}(\alpha), \ldots, c_{x_k}(\alpha))\} \in U\\
&\overset{\yaref{thm:los}}{\iff}&
(\cF, \tilde{\in }) \models \phi(\overline{j}(x_1), \ldots, \overline{j}(x_k)).
\end{IEEEeqnarray*}
Let us show that $(\cF, \tilde{\in })$
is well-founded.
Otherwise there is $\langle f_n : n < \omega \rangle$
such that $f_n \in {}^\kappa V$
and $[f_{n+1}] \tilde{\in } [f_n]$ for all $n < \omega$.
Then $X_n \coloneqq \{\xi < \kappa : f_{n+1}(\xi) \in f_n(\xi)\} \in U$,
so $\bigcap X_n \in U$.
Let $\xi_0 \in \bigcap X_n$.
Then $f_0(\xi_0) \ni f_1(\xi_0) \ni f_2(\xi_0) \ni \ldots$ $\lightning$.
Note that $\tilde{\in }$ is set-like,
therefore by the \yaref{lem:mostowski}
there is some transitive $M$ with $(\cF, \tilde{\in }) \overset{\sigma}{\cong} (M, \in )$.
We can now define an elementary embedding $j\colon V \to M$
by $j \coloneqq \sigma \circ \overline{j}$.
It remains to show that $\alpha < \kappa \implies j(\alpha) = \alpha$.
This can be done by induction:
Fix $\alpha$. We already know $j(\alpha) \ge \alpha$.
Suppose $\beta \in j(\alpha)$.
Then $\beta = \sigma([f])$ for some $f$
and $\sigma([f]) \in \sigma([c_{\alpha}])$,
i.e.~$[f] \tilde{\in} [c_\alpha]$.
Thus $\{\xi < \kappa : f(\xi) \in \underbrace{c_{\alpha}(\xi)}_{\alpha}\} \in U$.
Hence there is some $\delta < \alpha$
such that
\[
X_\delta \coloneqq \{\xi < \kappa : f(\xi) = \delta\} \in U,
\]
as otherwise $\forall \delta < \alpha. ~ \kappa \setminus X_\delta \in U$,
i.e.~$\emptyset = (\bigcap_{\delta < \alpha} \kappa \setminus X_{\delta}) \cap X \in U \lightning$.
We get $[f] = [c_{\delta}]$,
so $\beta = \sigma([f]) = \delta([c_{\delta}]) = j(\delta) = \delta$,
where for the last equality we have applied the induction hypothesis.
So $j(\alpha) \le \alpha$.
% It is also easy to show $j(\kappa) > \kappa$.
\end{proof}
\begin{theorem}[\L o\'s]
\yalabel{\L o\'s's Theorem}{\L o\'s}{thm:los}
For all formulae $\phi$ and for all $f_1, \ldots, f_k \in {}^{\kappa} V$,
\[
(\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
\iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.
\]
\end{theorem}
\begin{proof}
Induction on the complexity of $\phi$.
\end{proof}