fixed numbers of lectures and failed merge

inputs/lecture_07.tex

@@ -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 $\alpha \subseteq \beta$,

inputs/lecture_09.tex

@@ -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 $x \in A$,

inputs/lecture_10.tex

@@ -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 $A$,

inputs/lecture_16.tex

@@ -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

inputs/lecture_17.tex

@@ -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 $2^{\lambda} \ge \lambda^+$,
+    The hypothesis of \yaref{thm:silver}
-    strongly inaccessible cardinals are weakly inaccessible.
+    is consistent with $\ZFC$.
-
-    If $\GCH$ holds, the notions coincide.
 \end{remark}
 
-\begin{theorem}
+We will only proof \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$.
-    If $\kappa$ is inaccessible,
+The general proof differs only in notation.
-    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]
+\begin{remark}
-    A cardinal $\kappa > \aleph_0$ is \vocab{measurable} 
+    It is important that the cofinality is uncountable.
-    iff there is an ultrafilter $U$ on $\kappa$,
+    For example it is consistent
-    such that $U$ is not principal\footnote{%
+    with $\ZFC$ that
-        i.e.~$\{\xi\}  \not\in U$ for all $\xi < \kappa$%
+    $2^{\aleph_n} = \aleph_{n+1}$ for all $n < \omega$
-    }
+    but at the same time $2^{\aleph_{\omega}} = \aleph_{\omega + 2}$.
-    and
+\end{remark}
-    if $\theta < \kappa$
+
-    and $\{X_i : i < \theta\} \subseteq  U$,
+\begin{refproof}{thm:silver}
-    then $\bigcap_{i < \theta} X_i \in U$
+    We need to count the number of $X \subseteq \aleph_{\omega_1}.$
-\end{definition}
+    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{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 $\alpha$,
+        For $X,Y \subseteq \aleph_{\omega_1}$
-                $j(\alpha)$ is an ordinal such that $j(\alpha) \ge \alpha$.
+        it is $X \neq Y \iff f_X \neq f_Y$.
     \end{claim}
     \begin{subproof}
-                $\alpha \in \Ord$
+        $X \neq Y$ holds iff $X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha$
-                can be written as
+        for some $\alpha < \omega_1$.
-                \[\forall  x \in \alpha .~\forall y \in  x.~y \in \alpha
+        But then $f_X(\alpha) \neq f_Y(\alpha)$.
-                    \land \forall x \in \alpha .~ \forall y \in  \alpha.~(x \in y \lor x = y \lor y \in x).
-                \]
-                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 $j(\kappa) \neq \kappa$,
+    For $X, Y \subseteq \aleph_{\omega_1}$
-            we have $j(\kappa) > \kappa$,
+    write $X \le Y$ iff
-            i.e.~$\kappa \in j(\kappa)$.
+    \[
-        \item $U$ is an ultrafilter:
+    \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}
-            Let $X \subseteq  \kappa$.
+    \]
-            Then $\kappa \in j(\kappa) = j(X \cup (\kappa \setminus X)) = j(X) \cup j(\kappa \setminus X)$.
+    is stationary.
-            So $X \in U$ or $\kappa \setminus X \in U$.
 
-            Let $\theta < \kappa$
+    \begin{claim}
-            and $\{X_i : i < \theta\} \subseteq U$.
+        For all $X,Y \subseteq \aleph_{\omega_1}$,
-            Then $\kappa \in j(X_i)$ for all $i < \theta$,
+        $X \le Y$ 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
         \[
-            \kappa \in  \bigcap_{i < \theta} j(X_i)
+        f_{\aleph_\alpha}(Y \cap \aleph_\alpha) = f_Y(\alpha) = f_{Y'}(\alpha) = f_{\aleph_\alpha}(Y' \cap \aleph_\alpha),
-            = j\left( \bigcap_{i < \theta} X_i \right) \in U.
         \]
-            This holds since $j(\theta) = \theta$ (as $\theta < \kappa$),
+        i.e.~$Y \cap \aleph_\alpha = Y' \cap \aleph_\alpha$.
-            so $j(\langle X_i : i < \theta \rangle) = \langle j(X_i) : i < \theta \rangle$.
+        Since $T$ is cofinal in $\omega_1$,
+        it follows that $Y = Y'$.
+        So  $|A_3| \le 1 \lightning$
+    \end{subproof}
 
-            Also if $\xi < \kappa$,
+    Let us now define a sequence $\langle X_i : i < \aleph_{\omega_1 + 1} \rangle$
-            then $j(\{\xi\}) = \{\xi\}$
+    of subsets of $\aleph_{\omega_1 + 1}$ as follows:
-            so $\kappa \not\in  j(\{\xi\})$
-            and $\{\xi\}  \not\in U$.
-    \end{itemize}
 
-
+    Suppose $\langle X_j : j < i \rangle$
-    1. $\implies$ 2.
+    were already chosen.
-    Fix $U$.
+    Consider
-    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 $f_1, \ldots, f_k \in {}^{\kappa} V$,
     \[
-        (\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
+    \{Y \subseteq  \aleph_{\omega_1} : \exists j < i.~Y \le X_j\}
-        \iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.
+    = \bigcup_{j < i} \{Y \subseteq \aleph_{\omega_1} : Y \le X_j\}.
     \]
-\end{theorem}
+    This set has cardinality $\le \aleph_{\omega_1}$
-\begin{proof}
+    by \yaref{thm:silver:p:c3}.
-    Induction on the complexity of $\phi$.
+    Let $X_i \subseteq \aleph_{\omega_1}$
-\end{proof}
+    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}

inputs/lecture_18.tex

@@ -0,0 +1,239 @@
+\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}