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Josia Pietsch 2024-01-25 20:18:46 +01:00
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@ -1,4 +1,5 @@
\lecture{13}{2023-11-08}{} \lecture{13}{2023-11-08}{}
\gist{% \gist{%
% Recap % Recap
$\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $. $\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
@ -124,7 +125,8 @@ with $(f^{-1}(\{1\}), <)$.
and $f$ is continuous. and $f$ is continuous.
\end{proof} \end{proof}
% TODO: new section? \subsection{$\Pi_1^1$-ranks}
\gist{% \gist{%
Recall that a \vocab{rank} on a set $C$ Recall that a \vocab{rank} on a set $C$
is a map $\phi\colon C \to \Ord$. is a map $\phi\colon C \to \Ord$.

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@ -1,47 +1,52 @@
\lecture{14}{2023-12-01}{} \lecture{14}{2023-12-01}{}
% TODO ANKI-MARKER
\begin{theorem}[Moschovakis] \begin{theorem}[Moschovakis]
If $C$ is coanalytic, If $C$ is coanalytic,
then there exists a $\Pi^1_1$-rank on $C$. then there exists a $\Pi^1_1$-rank on $C$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
\gist{%
Pick a $\Pi^1_1$-complete set. Pick a $\Pi^1_1$-complete set.
It suffices to show that there is a rank on it. It suffices to show that there is a rank on it.
Then use the reduction to transfer Then use the reduction to transfer
it to any coanalytic set, it to any coanalytic set,
i.e.~for $x,y \in C'$ i.e.~for $x,y \in C'$
let let
\[ \[
x \le^{\ast}_{C'} y :\iff f(x) \le^\ast_{C} f(y) x \le^{\ast}_{C'} y :\iff f(x) \le^\ast_{C} f(y)
\] \]
and similarly for $<^\ast$. and similarly for $<^\ast$.
% https://q.uiver.app/#q=WzAsNSxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiQyJdLFsyLDIsIlxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d % https://q.uiver.app/#q=WzAsNCxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiXFxiZWdpbnthcnJheX17Y31DXFxcXFxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX1cXGVuZHthcnJheX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d
\[\begin{tikzcd} \begin{tikzcd}
Y && X \\ Y && X \\
{C'} && C \\ {C'} && {\begin{array}{c}C\\\Pi_1^1-\text{complete}\end{array}}
&& {\Pi_1^1-\text{complete}} \arrow["f", from=1-1, to=1-3]
\arrow["f", from=1-1, to=1-3] \arrow["\subseteq", hook, from=2-1, to=1-1]
\arrow["\subseteq", hook, from=2-1, to=1-1] \arrow["\subseteq"', hook, from=2-3, to=1-3]
\arrow["\subseteq"', hook, from=2-3, to=1-3] \end{tikzcd}
\end{tikzcd}\]
Let $X = 2^{\Q} \supseteq \WO$. Let $X = 2^{\Q} \supseteq \WO$.
We have already show that $\WO$ is $\Pi^1_1$-complete. We have already show that $\WO$ is $\Pi^1_1$-complete.
Set $\phi(x) \coloneqq \otp(x)$ Set $\phi(x) \coloneqq \otp(x)$
($\otp\colon \WO \to \Ord$ denotes the order type). ($\otp\colon \WO \to \Ord$ denotes the order type).
We show that this is a $\Pi^1_1$-rank. We show that this is a $\Pi^1_1$-rank.
}{%
It suffices to show this for a $\Pi^1_1$-complete set.
We show that $\phi \coloneqq \otp$ is a
$\Pi^1_1$-rank for $\WO \subseteq 2^{\Q}$.
}
Define $E \subseteq \Q^{\Q} \times 2^{\Q} \times 2^{\Q}$ Define $E \subseteq \Q^{\Q} \times 2^{\Q} \times 2^{\Q}$
by by
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{Cl}
(f,x,y) \in E &:\iff& f \text{ order embeds $(x, \le_{\Q})$ to $(y,\le _{\Q})$}\\ & (f,x,y) \in E\\
&\iff& \forall p,q \in \Q.~(p,q \in x \land p <_{\Q} q \implies f(p), f(q) \in y \land f(p) <_{\Q} f(q)) :\iff& f \text{ order embeds $(x, \le_{\Q})$ to $(y,\le _{\Q})$}\\
\iff& \forall p,q \in \Q.~(p,q \in x \land p <_{\Q} q \implies f(p), f(q) \in y \land f(p) <_{\Q} f(q))
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
$E$ is Borel as a countable intersection of clopen sets. $E$ is Borel\gist{ as a countable intersection of clopen sets}{}.
Define Define
$x <^\ast_{\phi}$ $x <^\ast_{\phi} y$
iff iff
\begin{itemize} \begin{itemize}
\item $(x, <_{\Q})$ is well ordered and \item $(x, <_{\Q})$ is well ordered and
@ -51,8 +56,9 @@
This is equivalent to This is equivalent to
\begin{itemize} \begin{itemize}
\item $x \in \WO$ and \item $x \in \WO$ and
\item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$. \item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$,
\end{itemize} \end{itemize}
so it is $\Pi^1_1$.
Furthermore $x \le_\phi^\ast y \iff$ Furthermore $x \le_\phi^\ast y \iff$
either $x <^\ast_\phi y$ or either $x <^\ast_\phi y$ or
@ -61,12 +67,11 @@
i.e.~either $x<^\ast_\phi y$ or i.e.~either $x<^\ast_\phi y$ or
$x,y \in \WO$ and any order embedding of $(x,<_{\Q})$ to $x,y \in \WO$ and any order embedding of $(x,<_{\Q})$ to
$(y, <_{\Q})$ is cofinal% $(y, <_{\Q})$ is cofinal%
\footnote{% \gist{\footnote{%
Recall that $A \subseteq (x,<_{\Q})$ Recall that $A \subseteq (x,<_{\Q})$
is \vocab{cofinal} if $\forall t \in x.~\exists a \in A.~t\le _{\Q} a$.% is \vocab{cofinal} if $\forall t \in x.~\exists a \in A.~t\le _{\Q} a$.%
} }}{}
in $(y, <_\Q)$ is in $(y, <_\Q)$ and vice versa.
cofinal in $(y, <_{\Q})$ and vice versa.
Equivalently, either $(x <^\ast_\phi y)$ Equivalently, either $(x <^\ast_\phi y)$
or or
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
@ -76,7 +81,6 @@
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\end{proof} \end{proof}
\begin{theorem} \begin{theorem}
\label{thm:uniformization} \label{thm:uniformization}
Let $X$ be Polish and $R \subseteq X \times \N$ by $\Pi^1_1$ Let $X$ be Polish and $R \subseteq X \times \N$ by $\Pi^1_1$
@ -88,7 +92,6 @@
\] \]
We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.% We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.%
\footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}} \footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}}
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
Let $\phi\colon R \to \Ord$ Let $\phi\colon R \to \Ord$
@ -112,27 +115,28 @@
and and
\[ \[
\bigcup_{n \in \N} C_n^\ast = \bigcup_{n \in \N} C_n. \bigcup_{n \in \N} C_n^\ast = \bigcup_{n \in \N} C_n.
\] \]
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Define $R \subseteq X \times \N$ by setting Define $R \subseteq X \times \N$ by setting
$(x,n) \in R :\iff x \in C_n$ $(x,n) \in R :\iff x \in C_n$
and apply \yaref{thm:uniformization}. and apply \yaref{thm:uniformization}.
\end{proof} \end{proof}
Let $X$ be a Polish space. Let $X$ be a Polish space.
If $(X, \prec)$ is well founded (i.e.~there are no infinite descending chains) If $(X, \prec)$ is well-founded%
then we define a rank $\rho_{y}\colon X > \Ord$ \gist{ (i.e.~there are no infinite descending chains)}{}
as follows: then we define a rank $\rho_{y}\colon X \to \Ord$ as follows:
For minimal elements the rank is $0$. For minimal elements the rank is $0$.
Otherwise set $\rho_<(x) \coloneqq \sup \{\rho_<(y) + 1 : y \prec x\}$. Otherwise set $\rho_\prec(x) \coloneqq \sup \{\rho_\prec(y) + 1 : y \prec x\}$.
Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$. Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
\begin{exercise} % \begin{exercise}
$\rho(\prec) \le |X|^+$ (successor cardinal). % $\rho(\prec) \le |X|^+$ (successor cardinal).
(for countable $<$) % (for countable $<$)
\todo{TODO} % \todo{TODO}
\end{exercise} % % TODO QUESTION
% \end{exercise}
\begin{theorem}[Kunen-Martin] \begin{theorem}[Kunen-Martin]
\yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin} \yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin}
If $(X, \prec)$ is wellfounded If $(X, \prec)$ is wellfounded
@ -140,80 +144,85 @@ Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
then $\rho(\prec) < \omega_1$. then $\rho(\prec) < \omega_1$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
% TODO GIST
% TODO QUESTION where did we use analytic?
Wlog.~$X = \cN$. Wlog.~$X = \cN$.
There is a tree $S$ on $\N \times \N \times \N$ There is a tree $S$ on $\N \times \N \times \N$
(i.e.~$S \subseteq \cN^3$) (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$)
such that such that
\[ \[
\forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \N.~(x,\alpha,y) \in [S]\right). \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right).
\] \]
Let Let
\[ \[
W \coloneqq \{w = (s_0,u_1,s_1,\ldots, u_n, s_n) : s_i, u_i \in \N^{n} W \coloneqq \{w = (s_0,u_1,s_1,\ldots, u_n, s_n) : s_i, u_i \in \N^{n}
\land (s_{i-1}, u_i, s_i) \in S\}. \land (s_{i-1}, u_i, s_i) \in S\}.
\] \]
So $|W| \le \aleph_0$. Clearly $|W| \le \aleph_0$.
Define $\prec^\ast$ on $W$ Define $\prec^\ast$ on $W$
by setting by setting
\[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\] \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
\begin{itemize} \begin{itemize}
\item $n < m$, \item $n < m$ and
\item $\forall i \le n.~s_i \subsetneq s_i'$ and \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
\item $\forall i \le n.~u_i \subsetneq u_i'$. %\todo{$\subseteq$ or $\subsetneq$?} % TODO QUESTION
\end{itemize} \end{itemize}
\begin{claim} \begin{claim}
$\prec^\ast$ is well-founded. $\prec^\ast$ is well-founded.
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
If $w_n = (s_0^n, u_1^n, \ldots, u_n^n, s_n^n)$ \gist{%
was descending, If $w_n = (s_0^n, u_1^n, \ldots, u_n^n, s_n^n)$
then let was descending,
\[ then let
x_i \coloneqq \bigcup s_i^n \in \cN \[
\] x_i \coloneqq \bigcup s_i^n \in \cN
and \]
\[ and
\alpha_i \coloneqq \bigcup_n u_i^n \cN. \[
\] \alpha_i \coloneqq \bigcup_n u_i^n \cN.
We get $(x_{i-1}, \alpha_i, x_i) \in [S]$, \]
hence $x_{i-1} \succ x_i$ for all $i$, We get $(x_{i-1}, \alpha_i, x_i) \in [S]$,
but this is an infinite descending chain hence $x_{i-1} \succ x_i$ for all $i$,
in the original relation $\lightning$ but this is an infinite descending chain
in the original relation $\lightning$
}{%
Use that $ \prec$ is well-founded.
}
\end{subproof} \end{subproof}
Hence $\rho(\prec^\ast) < |W|^+ \le \omega_1$.
% TODO QUESTION
We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$
with
\begin{IEEEeqnarray*}{rCl}
\rho(\prec) &=& \rho(T_{\prec})
\end{IEEEeqnarray*}
by setting $\emptyset \in T_{\prec}$
and
$(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$,
iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
\todo{Fix typos and end proof} For all $x \succ y$
% Hence $\rho(\prec^\ast) < \omega_1$. pick $\alpha_{x,y} \in \cN$
% such that $(x, \alpha_{x,y}, y) \in [S]$
% We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$ define
% with \begin{IEEEeqnarray*}{rCl}
% \begin{IEEEeqnarray*}{rCl} \phi\colon T_{\prec} \setminus \{\emptyset\} &\longrightarrow & W \\
% \rho(\prec) &=& \rho(T_{\prec}) \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots,
% \end{IEEEeqnarray*} \alpha_{x_{n-1}}, x_n\defon{n}).
% by setting $\emptyset \in T_{\prec}$ \end{IEEEeqnarray*}
% and Then $\phi$ is a homomorphism of $\subsetneq$ to $\prec^\ast$
% $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$, so
% iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$. \[
% \rho(\prec)
% For all $x \succ y$ = \rho(T_{\prec} \setminus \{\emptyset\} , \subsetneq)
% pick $\alpha_{x,y} \in \cN$ \le \rho(\prec^\ast)
% such that $(x, \alpha_{x,y}, y) \in [S]$ < \omega_1.
% define \]
% \begin{IEEEeqnarray*}{rCl}
% \phi\colon T_{\prec} &\longrightarrow & W \\
% \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots,
% \alpha_{x_{n-1}}, x_n\defon{n}).
% \end{IEEEeqnarray*}
% Then $\phi$ is a homomorphism of $\subsetneq$ to $<^\ast$
% and
% \[
% \rho(<) = \rho(T_{\prec}, \subsetneq) \le \rho(<^\ast) < \omega_1.
% \]
\todo{Exercise}
\end{proof} \end{proof}

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@ -1,5 +1,8 @@
\lecture{15}{2023-12-05}{} \lecture{15}{2023-12-05}{}
% TODO ANKI-MARKER
\begin{theorem}[Boundedness Theorem] \begin{theorem}[Boundedness Theorem]
\yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness} \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}