lecture 17

inputs/lecture_17.tex

@@ -0,0 +1,240 @@
+\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}

logic.sty

@@ -127,6 +127,7 @@
 
 
 \DeclareSimpleMathOperator{CH}
+\DeclareSimpleMathOperator{GCH}
 \DeclareSimpleMathOperator{DC}
 \DeclareSimpleMathOperator{Ord}
 \DeclareSimpleMathOperator{OR} % Ordinals

logic2.tex

@@ -40,6 +40,7 @@
 \input{inputs/lecture_14}
 \input{inputs/lecture_15}
 \input{inputs/lecture_16}
+\input{inputs/lecture_17}
 
 
 \cleardoublepage