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2
inputs/lecture_07.tex
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@ -9,7 +9,7 @@
\begin{enumerate}[(a)]
\item $0$ is an ordinal, and if $\alpha$ is
an ordinal, so is $\alpha + 1$.
\item If $\alpha$ is an ordinal and $x \in a$,
\item If $\alpha$ is an ordinal and $x \in \alpha$,
then $x$ is an ordinal.
\item If $\alpha, \beta$ are ordinals
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
such that for all $u,x$ there is exactly
one $y$ with $(u,x,y) \in D$
(basicalls $(u,x) \mapsto y$ is a function).
(basically $(u,x) \mapsto y$ is a function).
Then there is a unique function $f$ on $A$
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
\[
\rk_R \rank(R) \coloneqq \ran(F).
\rank(R) \coloneqq \ran(F).
\]
In the special case that $R$ is a linear order on $A$,

225
inputs/lecture_16.tex
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@ -1,4 +1,3 @@
<<<<<<< HEAD
\lecture{16}{2023-12-11}{}
Recall \yaref{thm:fodor}.
@ -238,227 +237,3 @@ then it is true at $\kappa$.
The proof of \yalabel{thm:silver} is quite elementary,
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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@ -1,240 +1,220 @@
\lecture{17}{2023-12-18}{Large cardinals}
\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{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.
The hypothesis of \yaref{thm:silver}
is consistent with $\ZFC$.
\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}
We will only proof \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$.
The general proof differs only in notation.
\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{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{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.
\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 $\lambda < \kappa$.
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 $[f], [g]$
For $X \subseteq \aleph_{\omega_1}$
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)).
f_X\colon \omega_1 &\longrightarrow & \aleph_{\omega_1} \\
\alpha &\longmapsto & f_{\aleph_\alpha}(X \cap \aleph_\alpha).
\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 $n < \omega$.
\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}
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 $f_0(\xi_0) \ni f_1(\xi_0) \ni f_2(\xi_0) \ni \ldots$ $\lightning$.
For $X, Y \subseteq \aleph_{\omega_1}$
write $X \le Y$ iff
\[
\{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}
\]
is stationary.
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 $\delta < \alpha$
\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
\[
X_\delta \coloneqq \{\xi < \kappa : f(\xi) = \delta\} \in U,
C \cap \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\} = \emptyset
\]
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 $f_1, \ldots, f_k \in {}^{\kappa} V$,
and
\[
(\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
\iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.
D \cap \{\alpha < \omega_1 : f_Y(\alpha) \le f_X(\alpha)\} = \emptyset.
\]
\end{theorem}
\begin{proof}
Induction on the complexity of $\phi$.
\end{proof}
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}

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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}