Josia Pietsch
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278 lines
12 KiB
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278 lines
12 KiB
TeX
\lecture{17}{2023-12-14}{Silver's Theorem}
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We now want to prove \yaref{thm:silver}.
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\gist{%
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\begin{remark}
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The hypothesis of \yaref{thm:silver}
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is consistent with $\ZFC$.
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\end{remark}
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}{}
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We will only prove
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\gist{%
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\yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$
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(see \yaref{thm:silver1}).
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The general proof differs only in notation.
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}{\yaref{thm:silver1}.}
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\gist{%
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\begin{remark}
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It is important that the cofinality is uncountable.
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For example it is consistent
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with $\ZFC$ that
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$2^{\aleph_n} = \aleph_{n+1}$ for all $n < \omega$
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but at the same time $2^{\aleph_{\omega}} = \aleph_{\omega + 2}$.
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\end{remark}
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}{}
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\begin{yarefproof}{thm:silver1}
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\gist{%
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We need to count the number of $X \subseteq \aleph_{\omega_1}.$
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Let us fix $\langle f_\lambda : \lambda < \kappa \text{ an infinite cardinal} \rangle$
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such that $f_{\lambda}\colon \cP(\lambda) \to \lambda^+$
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is bijective for each $\lambda < \kappa$.
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For $X \subseteq \aleph_{\omega_1}$
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define
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\begin{IEEEeqnarray*}{rCl}
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f_X\colon \omega_1 &\longrightarrow & \aleph_{\omega_1} \\
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\alpha &\longmapsto & f_{\aleph_\alpha}(X \cap \aleph_\alpha).
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\end{IEEEeqnarray*}
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\begin{claim}
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For $X,Y \subseteq \aleph_{\omega_1}$
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it is $X \neq Y \iff f_X \neq f_Y$.
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\end{claim}
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\begin{subproof}
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$X \neq Y$ holds iff $X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha$
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for some $\alpha < \omega_1$.
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But then $f_X(\alpha) \neq f_Y(\alpha)$.
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\end{subproof}
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For $X, Y \subseteq \aleph_{\omega_1}$
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write $X \le Y$ iff
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\[
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\{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}
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\]
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is stationary.
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\begin{claim}
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For all $X,Y \subseteq \aleph_{\omega_1}$,
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$X \le Y$ or $Y \le X$.
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\end{claim}
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\begin{subproof}
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Suppose that $X \nleq Y$ and $Y \nleq X$.
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Then there are clubs $C,D \subseteq \omega_1$
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such that
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\[
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C \cap \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\} = \emptyset
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\]
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and
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\[
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D \cap \{\alpha < \omega_1 : f_Y(\alpha) \le f_X(\alpha)\} = \emptyset.
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\]
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Note that $C \cap D$ is a club.
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Take some $\alpha \in C \cap D$.
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But then $f_X(\alpha) \le f_Y(\alpha)$
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or $f_{Y}(\alpha) \le f_X(\alpha)$ $\lightning$
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\end{subproof}
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\begin{claim}
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\label{thm:silver:p:c3}.
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Let $X \subseteq \aleph_{\omega_1}$.
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Then
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\[
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|\{Y \subseteq \aleph_{\omega_1} : Y \le X\}| \le \aleph_{\omega_1}.
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\]
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\end{claim}
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\begin{subproof}
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Write $A \coloneqq \{Y \subseteq X_{\omega_1} : Y \le X\}$.
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Suppose $|A| \ge \aleph_{\omega_1 + 1}$.
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For each $Y \in A$
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we have that
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\[
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S_Y \coloneqq \{\alpha : f_Y(\alpha) \le f_X(\alpha)\}
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\]
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is a stationary subset of $\omega_1$.
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Since by assumption $2^{\aleph_1} = \aleph_2$,
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there are at most $\aleph_2$ such $S_Y$.
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Suppose that for each $S \subseteq \omega_1$,
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\[
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|\{Y \in A : S_Y = S\}| < \aleph_{\omega_1 + 1}.
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\]
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Then $A$ is the union of $\le \aleph_2$ many
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sets of size $< \aleph_{\omega_1 + 1}$.
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Thus this is a contradiction since $\aleph_{\omega_1 + 1}$
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is regular.
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So there exists a stationary $S \subseteq \omega_1$
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such that
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\[
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A_1 = \{Y \subseteq \aleph_{\omega_1} : S_Y = S\}
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\]
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has cardinality $\aleph_{\omega_1 + 1}$.
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We have
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\[f_Y(\alpha) \le f_X(\alpha) = f_{\aleph_\alpha}(X \cap \aleph_\alpha)< \aleph_{\alpha + 1}\]
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for all $Y \in A_1, \alpha \in S$.
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Let $\langle g_{\alpha} : \alpha \in S \rangle$
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be such that $g_\alpha\colon \aleph_{\alpha} \twoheadrightarrow f_X(\alpha) + 1$
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is a surjection for all $\alpha \in S$.
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Then for each $Y \in A_1$ define
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\begin{IEEEeqnarray*}{rCl}
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\overline{f}_Y\colon S &\longrightarrow & \aleph_{\omega_1} \\
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\alpha &\longmapsto & \min \{\xi : g_\alpha(\xi) = f_Y(\alpha)\}.
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\end{IEEEeqnarray*}
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Let $D$ be the set of all limit ordinals $< \omega_1$.
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Then $S \cap D$ is a stationary set:
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If $C$ is a club, then $C \cap D$ is a club,
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hence $(S \cap D) \cap C = S \cap (D \cap C) \neq \emptyset$.
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Now to each $Y \in A$ we may associate
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a regressive function
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\begin{IEEEeqnarray*}{rCl}
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h_Y \colon S \cap D &\longrightarrow & \omega_1 \\
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\alpha &\longmapsto & \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_{\beta}\}.
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\end{IEEEeqnarray*}
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$h_Y$ is regressive, so by \yaref{thm:fodor}
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there is a stationary $T_Y \subseteq S \cap D$ on which $h_Y$ is constant.
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By an argument as before,
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there is a stationary $T \subseteq S \cap D$ such that
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\[
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|A_2| = \aleph_{\omega_1 +1},
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\]
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where $A_2 \coloneqq \{Y \in A_1 : T_Y = T\}$.
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Let $\beta < \omega_1$ be such that for all $Y \in A_2$
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and for all $\alpha \in T$, $h_Y(\alpha) = \beta$.
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% TODO WHY DOES THIS WORK?
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Then $\overline{f}_Y(\alpha) < \aleph_\beta$
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for all $Y \in A_2$ and $\alpha \in T$.
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There are at most $\aleph_\beta^{\aleph_1}$ many functions
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$T \to \aleph_\beta$,
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but
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\begin{IEEEeqnarray*}{rCl}
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\aleph_\beta^{\aleph_1} &\le & 2^{\aleph_\beta \cdot \aleph_1}\\
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&=& \aleph_{\beta+1} \cdot \aleph_2\\
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&<& \aleph_{\omega_1}.
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\end{IEEEeqnarray*}
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Suppose that for each function
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$\tilde{f}\colon T \to \aleph_\beta$
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there are $< \aleph_{\omega_1 + 1}$ many $Y \in A_2$
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with $\overline{f}_Y \cap T = \tilde{f}$.
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Then $A_2$ is the union of $<\aleph_{\omega_1}$
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many sets each of size $< \aleph_{\omega_1 + 1}$ $\lightning$.
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Hence for some $\tilde{f}\colon T \to \aleph_\beta$,
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\[
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|A_3| = \aleph_{\omega_1 + 1},
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\]
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where $A_3 = \{Y \in A_2 : \overline{f}_Y\defon{T} = \tilde{f}\}$.
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Let $Y, Y' \in A_3$ and $\alpha \in T$.
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Then
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\[
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\overline{f}_Y(\alpha) = \overline{f}_{Y'}(\alpha),
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\]
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hence
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\[
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f_{\aleph_\alpha}(Y \cap \aleph_\alpha) = f_Y(\alpha) = f_{Y'}(\alpha) = f_{\aleph_\alpha}(Y' \cap \aleph_\alpha),
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\]
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i.e.~$Y \cap \aleph_\alpha = Y' \cap \aleph_\alpha$.
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Since $T$ is cofinal in $\omega_1$,
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it follows that $Y = Y'$.
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So $|A_3| \le 1 \lightning$
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\end{subproof}
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Let us now define a sequence $\langle X_i : i < \aleph_{\omega_1 + 1} \rangle$
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of subsets of $\aleph_{\omega_1 + 1}$ as follows:
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Suppose $\langle X_j : j < i \rangle$
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were already chosen.
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Consider
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\[
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\{Y \subseteq \aleph_{\omega_1} : \exists j < i.~Y \le X_j\}
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= \bigcup_{j < i} \{Y \subseteq \aleph_{\omega_1} : Y \le X_j\}.
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\]
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This set has cardinality $\le \aleph_{\omega_1}$
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by \yaref{thm:silver:p:c3}.
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Let $X_i \subseteq \aleph_{\omega_1}$
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be such that $X_i \nleq X_j$ for all $j < i$.
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The set
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\[
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P \coloneqq \{Y \subseteq \aleph_{\omega_1} : \exists i < \aleph_{\omega_1 + 1} .~Y \le X_i\}
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= \bigcup_{i < \aleph_{\omega_1 + 1}} \{Y \subseteq \aleph_{\omega_1} : Y \le X_i\}
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\]
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has size $\le \aleph_{\omega_1 + 1}$
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(in fact the size is exactly $\aleph_{\omega_1 + 1}$).
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On the other hand
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$P = \cP(\aleph_{\omega_1})$
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because if $X \subseteq \aleph_{\omega_1}$
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is such that $X \nleq X_i$ for all $i < \aleph_{\omega_1 + 1}$,
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then $X_i \le X$ for all $i < \aleph_{\omega_1 + 1}$,
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so such a set $X$ does not exist by \yaref{thm:silver:p:c3}.
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}{Need to count $X \subseteq \aleph_{ \omega_{1}}$.
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\begin{itemize}
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\item Fix bijections $f_\lambda\colon 2^{\lambda} \to \lambda^+$
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for all infinite cardinals $\lambda < \kappa$.
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\item For $X \subseteq \aleph_{ \omega_1}$ define $f_X\colon \omega_1 \to \aleph_{ \omega_1},
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\alpha \mapsto f_{\aleph_\alpha}(X \cap \aleph_\alpha)$.
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\item (1) $X \neq Y \iff f_X \neq f_Y$:
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\begin{itemize}
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\item $X \neq Y \iff \exists \alpha.~X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha \iff \exists \alpha.~f_X(\alpha) \neq f_Y(\alpha)$.
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\end{itemize}
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\item $X \le Y :\iff \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}$ stationary.
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\item (2) $X\le Y \lor Y \le X$:
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Suppose $X \not\le Y, Y \not\le X$. Choose witnessing clubs $C, D$.
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$C \cap D$ is club, so $C\cap D \ni \alpha \implies f_X(\alpha) \substack{\nleq\\\ngeq} f_Y(\alpha) \lightning$
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\item (3) $X \subseteq \aleph_{ \omega_1}$, then $|\underbrace{\{Y \subseteq \aleph_{ \omega_1} : Y \le X\}}_{A}| \le \aleph_{ \omega_1}$
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\begin{itemize}
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\item Suppose $|A| \ge \aleph_{ \omega_1 + 1}$.
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\item $S_Y \coloneqq \{\alpha : f_Y(\alpha) \le f_X(\alpha)\}$ stationary for all $Y \in A$.
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$2^{\aleph_1} = \aleph_2 \implies$ at most $\aleph_2$ such $S_Y$.
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\item If $\forall S \in \cP( \omega_1).~ |\underbrace{\{Y \in A : S_Y = S\}}_{A_S}| < \aleph_{ \omega_1 + 1}$,
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then $|A| \le \aleph_2 \cdot <\aleph_{ \omega_1 + 1}$ $\lightning$ $\aleph_{ \omega_1 + 1}$ regular.
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\item So $\exists S \in \omega_1.~|A_S| = \aleph_{ \omega_1 + 1}$.
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\item Fix surjection $\langle g_\alpha : \aleph_\alpha \twoheadrightarrow f_X(\alpha) + 1 : \alpha \in S \rangle$.
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($f_Y(\alpha) \le f_X(\alpha) < \aleph_{\alpha+1}$)
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\item $\forall Y \in A_S$ define $\overline{f}_Y \colon S \to \aleph_{ \omega_1}, \alpha \mapsto \min \{\xi : g_\alpha(\xi) = f_Y(\alpha)\}$.
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\item $S^\ast$ ($S \cap$ limit ordinals) is stationary.
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\item $\forall Y \in A$ define $h_Y\colon S^\ast \to \omega_1, \alpha \mapsto \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_\beta\}$.
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\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.
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\item $\exists T.~|\underbrace{\{Y \in A_S : T_Y = T\}}_{A_{S,T}}| = \aleph_{ \omega_1 + 1}$.
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\item Let $\{\beta\} = h_Y''T$, i.e.~$\overline{f}_Y(\alpha) < \aleph_{\beta}$ for $Y\in A_{S,T}, \alpha \in T$.
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\item $\leftindex^T \aleph_\beta \le 2^{\aleph_\beta \cdot \aleph_1} = \aleph_{\beta+1} \aleph_2 < \aleph_{ \omega_1}$.
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\item $\exists \tilde{f}\colon T \to \aleph_\beta .~ |\underbrace{\{Y \in A_{S,T} : \overline{f}_Y\defon{T} = \tilde{f}\} }_{A_{S,T,\tilde{f}}}| = \aleph_{ \omega_1 + 1}$.
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\item $Y,Y' \in A_{S,T,\tilde{f}} \implies \forall \alpha \in T.~f_Y(\alpha) = f_{Y'}\left( \alpha \right) \overset{T \text{ unbounded}}{\implies} Y = Y'$
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Thus $|A_{S,T, \tilde{f}} | \le 1 \lightning$.
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\end{itemize}
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\item Define sequence $\langle X_i : i < \aleph_{ \omega_1 + 1} \rangle$
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of subsets of $\aleph_{\omega_1}$:
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\begin{itemize}
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\item Consider $\{Y \subseteq \aleph_{ \omega_1} : \exists j < i.~Y \le X_j\}$
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(cardinality $ \le \aleph_{ \omega_1}$),
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Take $X_i \subseteq \aleph_{ \omega_1}$
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such that $X_i \subseteq \aleph_{ \omega_1}$
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such that $X_i \not\le X_j$ for all $j < i$.
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\item $P \coloneqq \{Y \subseteq \aleph_{ \omega_1} : \exists i < \aleph_{ \omega_1 + 1}.~Y \le X_i\}$
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\end{itemize}
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\item $|P| \overset{\text{in fact } =}{\le} \aleph_{ \omega_1 + 1} \aleph_{ \omega_1} = \aleph_{ \omega_1 + 1}$ by (3).
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\item $X \in \cP(\aleph_{ \omega_1}) \setminus P \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
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$\lightning$ (3).
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Thus $P = \cP(\aleph_{ \omega_1})$.
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\end{itemize}
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}
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\end{yarefproof}
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