fixed numbers of lectures and failed merge

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

@@ -0,0 +1,220 @@
+\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{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}

inputs/lecture_18.tex

@@ -1,5 +1,4 @@
-\lecture{17}{2023-12-18}{Large cardinals}
+\lecture{18}{2023-12-18}{Large cardinals}
-
 
 \begin{definition}
     \begin{itemize}