\lecture{17}{2023-12-14}{Silver's Theorem}

We now want to prove \yaref{thm:silver}.

\gist{%
\begin{remark}
    The hypothesis of \yaref{thm:silver}
    is consistent with $\ZFC$.
\end{remark}
}{}

We will only prove
\gist{%
    \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$
    (see \yaref{thm:silver1}).
    The general proof differs only in notation.
}{\yaref{thm:silver1}.}
\gist{%
\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{yarefproof}{thm:silver1}
\gist{%
    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 $\alpha \in T$, $h_Y(\alpha) = \beta$.
        % TODO WHY DOES THIS WORK?
        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
    \[
    P \coloneqq \{Y \subseteq \aleph_{\omega_1} : \exists  i < \aleph_{\omega_1 + 1} .~Y \le X_i\}
    = \bigcup_{i < \aleph_{\omega_1 + 1}} \{Y \subseteq  \aleph_{\omega_1} : Y \le X_i\}
    \]
    has size $\le  \aleph_{\omega_1 + 1}$
    (in fact the size is exactly $\aleph_{\omega_1 + 1}$).

    On the other hand
    $P = \cP(\aleph_{\omega_1})$
    because if $X \subseteq \aleph_{\omega_1}$
    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}.
}{Need to count $X \subseteq \aleph_{ \omega_{1}}$.
    \begin{itemize}
        \item Fix bijections $f_\lambda\colon 2^{\lambda} \to \lambda^+$
            for all infinite cardinals $\lambda < \kappa$.
        \item For $X \subseteq \aleph_{ \omega_1}$ define $f_X\colon \omega_1 \to \aleph_{ \omega_1},
            \alpha \mapsto f_{\aleph_\alpha}(X \cap \aleph_\alpha)$.
        \item (1) $X \neq Y \iff f_X \neq f_Y$:
            \begin{itemize}
            \item $X \neq Y \iff \exists \alpha.~X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha \iff \exists \alpha.~f_X(\alpha) \neq  f_Y(\alpha)$.
            \end{itemize}
        \item $X \le Y :\iff \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}$ stationary.
        \item (2) $X\le Y \lor Y \le X$:
            Suppose $X \not\le Y, Y \not\le X$. Choose witnessing clubs $C, D$.
            $C \cap D$ is club, so $C\cap D \ni \alpha \implies f_X(\alpha) \substack{\nleq\\\ngeq} f_Y(\alpha) \lightning$
        \item (3) $X \subseteq \aleph_{ \omega_1}$, then $|\underbrace{\{Y \subseteq \aleph_{ \omega_1} : Y \le X\}}_{A}| \le \aleph_{ \omega_1}$
            \begin{itemize}
                \item Suppose $|A| \ge \aleph_{ \omega_1 + 1}$.
                \item $S_Y \coloneqq  \{\alpha : f_Y(\alpha) \le f_X(\alpha)\}$ stationary for all $Y \in A$.
                    $2^{\aleph_1} = \aleph_2 \implies$ at most $\aleph_2$ such $S_Y$.
                \item If $\forall  S \in \cP( \omega_1).~ |\underbrace{\{Y \in A : S_Y = S\}}_{A_S}| < \aleph_{ \omega_1 + 1}$,
                    then $|A| \le \aleph_2 \cdot  <\aleph_{ \omega_1 + 1}$ $\lightning$  $\aleph_{ \omega_1 + 1}$ regular.
                \item So $\exists S \in \omega_1.~|A_S| = \aleph_{ \omega_1 + 1}$.
                \item Fix surjection $\langle g_\alpha : \aleph_\alpha \twoheadrightarrow f_X(\alpha) + 1 : \alpha \in S \rangle$.
                    ($f_Y(\alpha) \le f_X(\alpha) < \aleph_{\alpha+1}$)
                \item $\forall  Y \in A_S$ define $\overline{f}_Y \colon  S \to  \aleph_{ \omega_1}, \alpha \mapsto \min \{\xi : g_\alpha(\xi) = f_Y(\alpha)\}$.
                \item $S^\ast$ ($S \cap$ limit ordinals) is stationary.
                \item $\forall Y \in A$ define $h_Y\colon S^\ast \to  \omega_1, \alpha \mapsto \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_\beta\}$.
                \item Apply \yaref{thm:fodor} to $h_Y, S^\ast$ to get $T_Y \subseteq S^\ast$ stationary with $h_Y\defon{T_Y}$ constant.
                \item $\exists T.~|\underbrace{\{Y \in A_S : T_Y = T\}}_{A_{S,T}}| = \aleph_{ \omega_1 + 1}$.
                \item Let $\{\beta\}  = h_Y''T$, i.e.~$\overline{f}_Y(\alpha) < \aleph_{\beta}$ for $Y\in A_{S,T}, \alpha \in T$.
                \item $\leftindex^T \aleph_\beta \le 2^{\aleph_\beta \cdot \aleph_1} = \aleph_{\beta+1} \aleph_2 < \aleph_{ \omega_1}$.
                \item $\exists \tilde{f}\colon T \to \aleph_\beta .~ |\underbrace{\{Y \in A_{S,T} : \overline{f}_Y\defon{T} = \tilde{f}\} }_{A_{S,T,\tilde{f}}}| = \aleph_{ \omega_1 + 1}$.
                \item $Y,Y' \in A_{S,T,\tilde{f}} \implies \forall \alpha \in T.~f_Y(\alpha) = f_{Y'}\left( \alpha \right) \overset{T \text{ unbounded}}{\implies} Y = Y'$
                    Thus $|A_{S,T, \tilde{f}} | \le 1 \lightning$.
            \end{itemize}

        \item Define sequence $\langle X_i : i < \aleph_{ \omega_1 + 1} \rangle$
            of subsets of $\aleph_{\omega_1}$:
            \begin{itemize}
                \item Consider $\{Y \subseteq \aleph_{ \omega_1} : \exists  j < i.~Y \le X_j\}$
                    (cardinality $ \le \aleph_{ \omega_1}$),
                    Take $X_i \subseteq \aleph_{ \omega_1}$
                    such that $X_i \subseteq \aleph_{ \omega_1}$
                    such that $X_i \not\le X_j$ for all $j < i$.
                \item $P \coloneqq \{Y \subseteq \aleph_{ \omega_1} : \exists  i < \aleph_{ \omega_1 + 1}.~Y \le X_i\}$
            \end{itemize}
        \item $|P| \overset{\text{in fact } =}{\le} \aleph_{ \omega_1 + 1} \aleph_{ \omega_1} = \aleph_{ \omega_1 + 1}$ by (3).
        \item $X \in \cP(\aleph_{ \omega_1}) \setminus P \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
            $\lightning$ (3).
            Thus $P = \cP(\aleph_{ \omega_1})$.
    \end{itemize}
}

\end{yarefproof}