diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index fb58f95..0a39244 100644 --- a/inputs/lecture_13.tex +++ b/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 $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$. diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex new file mode 100644 index 0000000..fec6d7e --- /dev/null +++ b/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} diff --git a/logic2.tex b/logic2.tex index 70c2c44..84f179a 100644 --- a/logic2.tex +++ b/logic2.tex @@ -39,6 +39,7 @@ \input{inputs/lecture_13} \input{inputs/lecture_14} \input{inputs/lecture_15} +\input{inputs/lecture_16} \cleardoublepage