w23-logic-2/inputs/lecture_17.tex

\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}