lecture 16

inputs/lecture_13.tex

@@ -130,6 +130,7 @@
 The next goal is to show the following:
 (However the method might be more interesting than the result)
 \begin{theorem}[Silver]
+    \yalabel{Silver's Theorem}{Silver}{thm:silver}
     If $2^{\aleph_\alpha} = \aleph_{\alpha + 1}$
     for all $\alpha < \omega_1$,
     then $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$.

inputs/lecture_16.tex

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

logic2.tex

@@ -39,6 +39,7 @@
 \input{inputs/lecture_13}
 \input{inputs/lecture_14}
 \input{inputs/lecture_15}
+\input{inputs/lecture_16}
 
 
 \cleardoublepage