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7 changed files with 51 additions and 67 deletions

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@ -39,6 +39,7 @@
is not a well-order on a countable set. is not a well-order on a countable set.
Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$. Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$.
\todo{move this}
\end{remark} \end{remark}
}{} }{}

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@ -155,15 +155,6 @@ We have shown (assuming \AxC to choose contained clubs):
= \bigcap_{\beta < \alpha} ([0,\beta] \cup A_\beta) = \bigcap_{\beta < \alpha} ([0,\beta] \cup A_\beta)
\] \]
\end{definition} \end{definition}
\begin{remark}+
\label{rem:diagiclosed}
Note that if $A$ is closed,
so is $[0,\alpha] \cup A$.
Since the intersection of arbitrarily many
closed sets is closed,
we get that the diagonal intersection
of closed sets is closed.
\end{remark}
\begin{lemma} \begin{lemma}
\label{lem:diagiclub} \label{lem:diagiclub}
Let $\kappa$ be a regular, uncountable cardinal. Let $\kappa$ be a regular, uncountable cardinal.
@ -190,23 +181,22 @@ We have shown (assuming \AxC to choose contained clubs):
$\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$. $\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$.
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
Cf.~\yaref{rem:diagiclosed}. Let $\gamma < \kappa$ be such that $\left( \diagi_{\beta < \kappa} D_{\beta} \right) \cap \gamma$
% Let $\gamma < \kappa$ be such that $\left( \diagi_{\beta < \kappa} D_{\beta} \right) \cap \gamma$ is unbounded in $\gamma$.
% is unbounded in $\gamma$. We aim to show that $\gamma \in \diagi_{\beta < \kappa} D_{\beta}$.
% We aim to show that $\gamma \in \diagi_{\beta < \kappa} D_{\beta}$. Let $\beta_0 < \gamma$.
% Let $\beta_0 < \gamma$. We need to see that $\gamma \in D_{\beta_0}$.
% We need to see that $\gamma \in D_{\beta_0}$.
% For each $\beta_0 \le \beta' < \gamma$ For each $\beta_0 \le \beta' < \gamma$
% there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$ there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$
% such that $\beta' \le \beta'' < \gamma$, such that $\beta' \le \beta'' < \gamma$,
% since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$. since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$.
% In particular $\beta'' \in D_{\beta_0}$. In particular $\beta'' \in D_{\beta_0}$.
% So $D_{\beta_0} \cap \gamma$ So $D_{\beta_0} \cap \gamma$
% is unbounded in $\gamma$. is unbounded in $\gamma$.
% Since $D_{\beta_0}$ is closed Since $D_{\beta_0}$ is closed
% it follows that $\gamma \in D_{\beta_0}$. it follows that $\gamma \in D_{\beta_0}$.
%As $\beta_0 < \gamma$ was arbitrary, %As $\beta_0 < \gamma$ was arbitrary,
%this shows that $\gamma \in \diagi_{\beta < n} D_\beta$. %this shows that $\gamma \in \diagi_{\beta < n} D_\beta$.
@ -280,9 +270,9 @@ We have shown (assuming \AxC to choose contained clubs):
\end{refproof} \end{refproof}
\begin{remark}+ \begin{remark}+
$\diagi_{\beta < \kappa} C_{\beta}$ actually $\diagi_{\beta < \kappa} C_{\beta}$ actually
\emph{is} a club, \emph{is} a club:
since $\diagi_{\beta < \kappa} C_\beta$ is closed, It suffices to show that $\diagi_{\beta < \kappa} C_\beta$ is closed.
again cf.~\yaref{rem:diagiclosed}. This can be shown in the same way as for $\diagi_{\beta < \kappa} D_\beta$.
% Let $\lambda < \kappa$ be a limit ordinal. % Let $\lambda < \kappa$ be a limit ordinal.
% Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$. % Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
% Then there exists $\alpha < \lambda$ such that % Then there exists $\alpha < \lambda$ such that
@ -298,20 +288,6 @@ We have shown (assuming \AxC to choose contained clubs):
iff $C \cap S \neq \emptyset$ iff $C \cap S \neq \emptyset$
for every club $C \subseteq \kappa$. for every club $C \subseteq \kappa$.
\end{definition} \end{definition}
\begin{remark}+[\url{https://mathoverflow.net/q/37503}]
Informally, club sets and stationary sets
can be viewed as large sets of a measure space
of measure $1$.
Clubs behave similarly to sets of measure $1$
and stationary sets are analogous to
sets of positive measure:
\begin{itemize}
\item Every club is stationary,
\item the intersection of two clubs is a club,
\item the intersection of a club and a stationary set is stationary,
\item there exist disjoint stationary sets.
\end{itemize}
\end{remark}
\begin{example} \begin{example}
\begin{itemize} \begin{itemize}
\item Every $D \subseteq \kappa$ which is club in $\kappa$ \item Every $D \subseteq \kappa$ which is club in $\kappa$

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@ -43,6 +43,7 @@ Recall that $F \subseteq \cP(\kappa)$ is a filter if
$X,Y \in F \implies X \cap Y \in F$, $X,Y \in F \implies X \cap Y \in F$,
$X \in F, X \subseteq Y \subseteq \kappa \implies Y \in F$ $X \in F, X \subseteq Y \subseteq \kappa \implies Y \in F$
and $\emptyset \not\in F, \kappa \in F$. and $\emptyset \not\in F, \kappa \in F$.
\todo{Move this to the definition of filter?}
}{} }{}
\begin{definition} \begin{definition}
A filter $F$ is an \vocab{ultrafilter} A filter $F$ is an \vocab{ultrafilter}
@ -103,7 +104,11 @@ one cofinality.
\begin{refproof}{thm:solovay}% \begin{refproof}{thm:solovay}%
\gist{% \gist{%
\footnote{``This is one of the arguments where it is certainly worth it to look at it again.''} %\footnote{``This is one of the arguments where it is certainly
% worth it to look at it again''}
% TODO: Look at this again and think about it.
% TODO TODO TODO
We will only prove this for $\aleph_1$. We will only prove this for $\aleph_1$.
Fix $S \subseteq \aleph_1$ stationary. Fix $S \subseteq \aleph_1$ stationary.
@ -192,22 +197,19 @@ one cofinality.
Write $\delta_i = \delta'$ and $T_i = T$. Write $\delta_i = \delta'$ and $T_i = T$.
By construction, all the $T_i$ are stationary. \begin{claim}
Since the $\delta_i$ are strictly increasing \label{thm:solovay:p:c2}
and since $\gamma_n^{\alpha} = \delta_i$ for all $\alpha \in T_i$, Each $T_i$ is stationary
we have that the $T_i$ are disjoint. and if $i \neq j$, then $T_i \cap T_j = \emptyset$.
% \begin{claim} \footnote{maybe this should not be a claim}
% \label{thm:solovay:p:c2} \end{claim}
% Each $T_i$ is stationary \begin{subproof}
% and if $i \neq j$, then $T_i \cap T_j = \emptyset$. The first part is true by construction.
% \end{claim} Let $j < i$.
% \begin{subproof} Then if $\alpha \in T_i$, $\alpha' \in T_j$,
% The first part is true by construction. we get $\gamma_n^{\alpha'} = \delta_j < \delta_i = \gamma_n^{\alpha}$
% Let $j < i$. hence $\alpha \neq \alpha'$.
% Then if $\alpha \in T_i$, $\alpha' \in T_j$, \end{subproof}
% we get $\gamma_n^{\alpha'} = \delta_j < \delta_i = \gamma_n^{\alpha}$
% hence $\alpha \neq \alpha'$.
% \end{subproof}
Now let Now let
\[ \[
@ -230,6 +232,8 @@ one cofinality.
\item $\exists n < \omega.~\forall \delta < \omega_1: \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\}$ \item $\exists n < \omega.~\forall \delta < \omega_1: \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\}$
stationary: stationary:
\begin{itemize} \begin{itemize}
% TODO THINK!
% TODO TODO TODO
\item Otherwise $\forall n < \omega.~\exists \delta.~\{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\} $ nonstationary. \item Otherwise $\forall n < \omega.~\exists \delta.~\{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\} $ nonstationary.
\item $\delta_n\coloneqq $ least such $\delta$, \item $\delta_n\coloneqq $ least such $\delta$,
$C_n$ club s.t.~$C_n \cap \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta_n\} = \emptyset$. $C_n$ club s.t.~$C_n \cap \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta_n\} = \emptyset$.

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@ -1,6 +1,9 @@
\lecture{17}{2023-12-14}{Silver's Theorem} \lecture{17}{2023-12-14}{Silver's Theorem}
We now want to prove \yaref{thm:silver}. 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 $2^{\kappa} = \kappa^+$.
\gist{% \gist{%
\begin{remark} \begin{remark}
@ -9,7 +12,7 @@ We now want to prove \yaref{thm:silver}.
\end{remark} \end{remark}
}{} }{}
We will only prove We will only proof
\gist{% \gist{%
\yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$ \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$
(see \yaref{thm:silver1}). (see \yaref{thm:silver1}).
@ -150,7 +153,6 @@ We will only prove
Let $\beta < \omega_1$ be such that for all $Y \in A_2$ Let $\beta < \omega_1$ be such that for all $Y \in A_2$
and for all $\alpha \in T$, $h_Y(\alpha) = \beta$. and for all $\alpha \in T$, $h_Y(\alpha) = \beta$.
% TODO WHY DOES THIS WORK?
Then $\overline{f}_Y(\alpha) < \aleph_\beta$ Then $\overline{f}_Y(\alpha) < \aleph_\beta$
for all $Y \in A_2$ and $\alpha \in T$. for all $Y \in A_2$ and $\alpha \in T$.

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@ -27,6 +27,7 @@
Since $2^{\lambda} < \kappa$, Since $2^{\lambda} < \kappa$,
\AxPow works. \AxPow works.
The other axioms are trivial. The other axioms are trivial.
\todo{Exercise}
\end{proof} \end{proof}
\begin{corollary} \begin{corollary}
$\ZFC$ does not prove the existence of inaccessible $\ZFC$ does not prove the existence of inaccessible

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@ -79,7 +79,6 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
A similar arguments yields \vocab{upwards absoluteness} for $\Sigma_1$-formulas A similar arguments yields \vocab{upwards absoluteness} for $\Sigma_1$-formulas
and \vocab{downwards absoluteness} for $\Pi_1$-formulas: and \vocab{downwards absoluteness} for $\Pi_1$-formulas:
\begin{lemma} \begin{lemma}
\label{lem:pi1downardsabsolute}
Let $M$ be transitive. Let $M$ be transitive.
Let $\phi(x_0,\ldots,x_n) \in \cL_\in$ and $a_0,\ldots,a_n \in M$. Let $\phi(x_0,\ldots,x_n) \in \cL_\in$ and $a_0,\ldots,a_n \in M$.
Then Then

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@ -112,7 +112,7 @@ is well founded.
The formula $\forall x.~\forall y .~((\forall z \in x.~z \in y \land \forall z \in y.~z \in x) \to x = y)$ The formula $\forall x.~\forall y .~((\forall z \in x.~z \in y \land \forall z \in y.~z \in x) \to x = y)$
is $\Pi_1$, hence it is true in $M[g]$ is $\Pi_1$, hence it is true in $M[g]$
by \yaref{lem:pi1downardsabsolute}. by %TODO REF downward absolutenes.
\item \AxFund: \item \AxFund:
Again, Again,
\[ \[
@ -137,7 +137,7 @@ is well founded.
so $M[g] \models \text{``$\{x,y\}$ is the pair of $x$ and $y$''}$. so $M[g] \models \text{``$\{x,y\}$ is the pair of $x$ and $y$''}$.
Hence $M[g] \models \AxPair$. Hence $M[g] \models \AxPair$.
\item \AxUnion: \item \AxUnion:
Similar to \AxPair. Similar to \AxPair.\gist{\todo{Exercise}}{}
\end{itemize} \end{itemize}
\end{proof} \end{proof}
@ -194,6 +194,7 @@ Still missing are
\[ \[
p \Vdash^{\mathbb{P}}_M \phi(\tau_1,\ldots, \tau_k). p \Vdash^{\mathbb{P}}_M \phi(\tau_1,\ldots, \tau_k).
\] \]
\end{enumerate} \end{enumerate}
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}