\lecture{18}{2023-12-18}{Large cardinals} \begin{definition} \label{def:inaccessible} \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 $(V_{\kappa}, \in\defon{V_\kappa}) \models\ZFC$.} \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\gist{\footnote{% i.e.~$\{\xi\} \not\in U$ for all $\xi < \kappa$% }}{} and $< \kappa$-closed\gist{,% i.e.~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.: Fix $j\colon V \to M$. 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} $\alpha \in \OR$ 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 $\leftindex^{\kappa}V$ be the class of all function from $\kappa$ to $V$. For $f,g \in \leftindex^{\kappa}V$ 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 \leftindex^\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 $\cF = \{[f] : f \in \leftindex^{\kappa}V\}$ 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 $f_n \in \leftindex^\kappa V$ 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 $f_1, \ldots, f_k \in \leftindex^{\kappa} V$, \[ (\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k]) \iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U. \] \end{theorem} \begin{proof} Induction on the complexity of $\phi$. \end{proof}