Compare commits
No commits in common. "ab14be172f2a77a6d74e679050918371f49cf94a" and "f4626be14bdd7767d155102625c0d4b5eaa1d9f8" have entirely different histories.
ab14be172f
...
f4626be14b
7 changed files with 51 additions and 67 deletions
|
@ -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}
|
||||||
}{}
|
}{}
|
||||||
|
|
||||||
|
|
|
@ -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 $\kappa$
|
||||||
|
|
|
@ -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.~$C_n \cap \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta_n\} = \emptyset$.
|
||||||
|
|
|
@ -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 $\alpha \in T$.
|
||||||
|
|
||||||
|
|
|
@ -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
|
||||||
|
|
|
@ -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
|
||||||
|
|
|
@ -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}
|
||||||
|
|
Loading…
Reference in a new issue