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lecture 17

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\lecture{17}{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}