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gist for lecture 13
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20
inputs/lecture_05.tex
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@ -103,7 +103,9 @@
Let $X$ be a completely metrizable space. Let $X$ be a completely metrizable space.
Then every comeager set of $X$ is dense in $X$. Then every comeager set of $X$ is dense in $X$.
\end{theorem} \end{theorem}
\todo{Proof (copy from some other lecture)} \gist{%
\todo{Proof (copy from some other lecture)}
}{Not proved in the lecture.}
\begin{theoremdef} \begin{theoremdef}
Let $X$ be a topological space. Let $X$ be a topological space.
The following are equivalent: The following are equivalent:
@ -118,7 +120,21 @@
\footnote{cf.~\yaref{s5e1}} \footnote{cf.~\yaref{s5e1}}
\end{theoremdef} \end{theoremdef}
\begin{proof} \begin{proof}
\todo{Proof (short)} (i) $\implies$ (ii)
\gist{%
Consider a comeager set $A$.
Let $U\neq \emptyset$ be any open set. Since $U$ is
non-meager, we have $A \cap U \neq \emptyset$.
}{The intersection of a comeager and a non-meager set is nonempty.}
(ii) $\implies$ (iii)
The complement of an open dense set is nwd.
\gist{%
Hence the intersection of countable
many open dense sets is comeager.
}{}
(iii) $\implies$ (i) (iii) $\implies$ (i)
Let us first show that $X$ is non-meager. Let us first show that $X$ is non-meager.

183
inputs/lecture_07.tex
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@ -77,7 +77,7 @@
sets that are not clopen)}{}. sets that are not clopen)}{}.
\end{example} \end{example}
\subsection{Turning Borels Sets into Clopens} \subsection{Turning Borel Sets into Clopens}
\begin{theorem}% \begin{theorem}%
\gist{% \gist{%
@ -109,55 +109,74 @@
into $B$. into $B$.
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Pick $\cT_B \supset \cT$ \gist{%
such that $(X, \cT_B)$ is Polish, Pick $\cT_B \supset \cT$
$B$ is clopen in $\cT_B$ and such that $(X, \cT_B)$ is Polish,
$\cB(X,\cT) = \cB(X, \cT_B)$. $B$ is clopen in $\cT_B$ and
$\cB(X,\cT) = \cB(X, \cT_B)$.
Therefore $(\cB, \cT_B\defon{B})$ is Polish. Therefore $(\cB, \cT_B\defon{B})$ is Polish.
We know that there is an embedding We know that there is an embedding
$f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$. $f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$.
Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$. Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$.
This is still continuous as $\cT \subseteq \cT_B$. This is still continuous as $\cT \subseteq \cT_B$.
Since $2^{\omega}$ is compact, $f$ is an embedding. Since $2^{\omega}$ is compact, $f$ is an embedding.
%\todo{Think about this} }{%
Clopenize $B$.
We can embed $2^{ \omega}$ into Polish spaces.
Clopenization makes the topology finer,
so this is still continuous wrt.~the original topology.
$2^{\omega}$ is compact, so this is an embedding.
}
\end{proof} \end{proof}
\begin{refproof}{thm:clopenize} \begin{refproof}{thm:clopenize}
We show that \gist{%
\begin{IEEEeqnarray*}{rCl} We show that
A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\ \begin{IEEEeqnarray*}{rCl}
&& (X, \cT_B) \text{ is Polish},\\ A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\
&& \cB(X, \cT) = \cB(X, \cT_B)\\ && (X, \cT_B) \text{ is Polish},\\
&& B \text{ is clopen in $\cT_B$}\\ && \cB(X, \cT) = \cB(X, \cT_B)\\
\} && B \text{ is clopen in $\cT_B$}\\
\end{IEEEeqnarray*} \}
is equal to the set of Borel sets. \end{IEEEeqnarray*}
is equal to the set of Borel sets.
The proof rests on two lemmata: }{%
Let $A$ be the set of clopenizable sets.
We show that $A = \cB(X)$.
}
\gist{The proof rests on two lemmata:}{}
\begin{lemma} \begin{lemma}
\label{thm:clopenize:l1} \label{thm:clopenize:l1}
Let $(X,\cT)$ be a Polish space. \gist{%
Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$) Let $(X,\cT)$ be a Polish space.
there is $\cT_F \supseteq \cT$ Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$)
such that $\cT_F$ is Polish, there is $\cT_F \supseteq \cT$
$\cB(\cT) = \cB(\cT_F)$ such that $\cT_F$ is Polish,
and $F$ is clopen in $\cT_F$. $\cB(\cT) = \cB(\cT_F)$
and $F$ is clopen in $\cT_F$.
}{%
Closed sets can be clopenized.
}
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$. \gist{%
Both are Polish spaces. Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
Take the coproduct% Both are Polish spaces.
\footnote{In the lecture, this was called the \vocab{topological sum}.} Take the coproduct%
$F \oplus (X \setminus F)$ of these spaces. \footnote{In the lecture, this was called the \vocab{topological sum}.}
This space is Polish, $F \oplus (X \setminus F)$ of these spaces.
and the topology is generated by $\cT \cup \{F\}$, This space is Polish,
hence we do not get any new Borel sets. and the topology is generated by $\cT \cup \{F\}$,
hence we do not get any new Borel sets.
}{Consider $(F, \cT\defon{F}) \oplus (X \setminus F, \cT\defon{X \setminus F})$.}
\end{proof} \end{proof}
So all closed sets are in $A$. \gist{%
Furthermore $A$ is closed under complements, So all closed sets are in $A$.
since complements of clopen sets are clopen. Furthermore $A$ is closed under complements,
since complements of clopen sets are clopen.
}{So $\Sigma^0_1(X), \Pi^0_1(X) \subseteq A$.}
\begin{lemma} \begin{lemma}
\label{thm:clopenize:l2} \label{thm:clopenize:l2}
@ -170,22 +189,23 @@
is Polish is Polish
and $\cB(\cT_\infty) = \cB(T)$. and $\cB(\cT_\infty) = \cB(T)$.
\end{lemma} \end{lemma}
\begin{proof} \begin{refproof}{thm:clopenize:l2}
We have that $\cT_\infty$ is the smallest \gist{%
topology containing all $\cT_n$. We have that $\cT_\infty$ is the smallest
To get $\cT_\infty$ topology containing all $\cT_n$.
consider To get $\cT_\infty$ consider
\[ \[
\cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}. \cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}.
\] \]
Then Then
\[ \[
\cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}. \cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}.
\] \]
(It suffices to take countable unions, (It suffices to take countable unions,
since we may assume that the $A_1, \ldots, A_n$ in the since we may assume that the $A_1, \ldots, A_n$ in the
definition of $\cF$ belong to definition of $\cF$ belong to
a countable basis of the respective $\cT_n$). a countable basis of the respective $\cT_n$).
}{}
% Proof was finished in lecture 8 % Proof was finished in lecture 8
Let $Y = \prod_{n \in \N} (X, \cT_n)$. Let $Y = \prod_{n \in \N} (X, \cT_n)$.
@ -195,6 +215,7 @@
\begin{claim} \begin{claim}
$\delta$ is a homeomorphism. $\delta$ is a homeomorphism.
\end{claim} \end{claim}
\gist{%
\begin{subproof} \begin{subproof}
Clearly $\delta$ is a bijection. Clearly $\delta$ is a bijection.
We need to show that it is continuous and open. We need to show that it is continuous and open.
@ -219,28 +240,34 @@
\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D. \delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.
\] \]
\end{subproof} \end{subproof}
}{}
This will finish the proof since \begin{claim}
\[ $D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y.$
D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y \end{claim}
\] \gist{%
Why? Let $(x_n) \in Y \setminus D$. \begin{subproof}
Then there are $i < j$ such that $x_i \neq x_j$. Let $(x_n) \in Y \setminus D$.
Take disjoint open $x_i \in U$, $x_j \in V$. Then there are $i < j$ such that $x_i \neq x_j$.
Then Take disjoint open $x_i \in U$, $x_j \in V$.
\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\] Then
is open in $Y\setminus D$. \[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]
Hence $Y \setminus D$ is open, thus $D$ is closed. is open in $Y\setminus D$.
It follows that $D$ is Polish. Hence $Y \setminus D$ is open, thus $D$ is closed.
\end{proof} \end{subproof}
It follows that $D$ is Polish.
}{}
\end{refproof}
We need to show that $A$ is closed under countable unions. \gist{%
By \yaref{thm:clopenize:l2} there exists a topology We need to show that $A$ is closed under countable unions.
$\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$ By \yaref{thm:clopenize:l2} there exists a topology
and $\cB(\cT_\infty) = \cB(\cT)$. $\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$
Applying \yaref{thm:clopenize:l1} and $\cB(\cT_\infty) = \cB(\cT)$.
yields a topology $\cT_\infty'$ such that Applying \yaref{thm:clopenize:l1}
$(X, \cT_\infty')$ is Polish, yields a topology $\cT_\infty'$ such that
$\cB(\cT_\infty') = \cB(\cT)$ $(X, \cT_\infty')$ is Polish,
and $A $ is clopen in $\cT_{\infty}'$. $\cB(\cT_\infty') = \cB(\cT)$
and $A $ is clopen in $\cT_{\infty}'$.
}{}
\end{refproof} \end{refproof}

20
inputs/lecture_08.tex
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@ -1,8 +1,9 @@
\lecture{08}{2023-11-10}{}\footnote{% \lecture{08}{2023-11-10}{}%
\gist{\footnote{%
In the beginning of the lecture, we finished In the beginning of the lecture, we finished
the proof of \yaref{thm:clopenize:l2}. the proof of \yaref{thm:clopenize:l2}.
This has been moved to the notes on lecture 7.% This has been moved to the notes on lecture 7.%
} }}{}
\subsection{Parametrizations} \subsection{Parametrizations}
%\todo{choose better title} %\todo{choose better title}
@ -22,13 +23,14 @@ where $X$ is a metrizable, usually second countable space.
\item $\{U_y : y \in Y\} = \Gamma(X)$. \item $\{U_y : y \in Y\} = \Gamma(X)$.
\end{itemize} \end{itemize}
\end{definition} \end{definition}
\gist{%
\begin{example} \begin{example}
Let $X = \omega^\omega$, $Y = 2^{\omega}$ Let $X = \omega^\omega$, $Y = 2^{\omega}$
and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$. and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$.
We will show that there is a $2^{\omega}$-universal We will show that there is a $2^{\omega}$-universal
set for $\Gamma$. set for $\Gamma$.
\end{example} \end{example}
}{}
\begin{theorem} \begin{theorem}
\label{thm:cantoruniversal} \label{thm:cantoruniversal}

5
inputs/lecture_11.tex
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@ -99,9 +99,10 @@
(continuous wrt.~to the topology of $X$) (continuous wrt.~to the topology of $X$)
On the other hand On the other hand
\[ \[
X \hookrightarrow\cN \overset{\text{continuous embedding}}{\hookrightarrow}\cC X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC
\] \]
\todo{second inclusion was on a homework sheet} \footnotetext{cf.~\yaref{s2e4}}
For the first inclusion, For the first inclusion,
recall that there is a continuous bijection $b\colon D \to X$, recall that there is a continuous bijection $b\colon D \to X$,
where $D \overset{\text{closed}}{\subseteq} \cN$. where $D \overset{\text{closed}}{\subseteq} \cN$.

39
inputs/lecture_13.tex
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@ -79,8 +79,6 @@ with $(f^{-1}(\{1\}), <)$.
}{easy} }{easy}
\end{proof} \end{proof}
% TODO ANKI-MARKER
\begin{theorem}[Lusin-Sierpinski] \begin{theorem}[Lusin-Sierpinski]
The set $\LO \setminus \WO$ The set $\LO \setminus \WO$
(resp.~$2^{\Q} \setminus \WO$) (resp.~$2^{\Q} \setminus \WO$)
@ -89,19 +87,23 @@ with $(f^{-1}(\{1\}), <)$.
\begin{proof} \begin{proof}
We will find a continuous function We will find a continuous function
$f\colon \Tr \to \LO$ such that $f\colon \Tr \to \LO$ such that
\gist{%
\[ \[
x \in \WF \iff f(x) \in \WO x \in \WF \iff f(x) \in \WO
\] \]
(equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$). (equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$).
This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete
}{
$f^{-1}(\LO \setminus \WO) = \IF$.
This suffices
}
(see \yaref{cor:ifs11c}). (see \yaref{cor:ifs11c}).
Fix a bijection $b\colon \N \to \N^{<\N}$. Fix a bijection $b\colon \N \to \N^{<\N}$.
\begin{idea} \begin{idea}
For $T \in \Tr$ consider For $T \in \Tr$ consider
$<_{KB}\defon{T}$ $<_{KB}\defon{T}$.
% TODO?
\end{idea} \end{idea}
Let $\alpha \in \Tr$. Let $\alpha \in \Tr$.
@ -109,7 +111,7 @@ with $(f^{-1}(\{1\}), <)$.
(i.e.~$m \le_{f(\alpha)} n$) (i.e.~$m \le_{f(\alpha)} n$)
iff iff
\begin{itemize} \begin{itemize}
\item $(\alpha(b(m)) = \alpha(b(n)) = 1$ \item $\alpha(b(m)) = \alpha(b(n)) = 1$
and $b(m) \le_{KB} b(n)$ and $b(m) \le_{KB} b(n)$
(recall that we identified $\Tr$ (recall that we identified $\Tr$
with a subset of ${2^{\N}}^{<\N}$), with a subset of ${2^{\N}}^{<\N}$),
@ -123,15 +125,16 @@ with $(f^{-1}(\{1\}), <)$.
\end{proof} \end{proof}
% TODO: new section? % TODO: new section?
\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 $\phi\colon C \to \Ord$.
\begin{example} \begin{example}
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\otp \colon \WO &\longrightarrow & \Ord \\ \otp \colon \WO &\longrightarrow & \Ord \\
x &\longmapsto & \text{the unique $\alpha \in \Ord$ such that $x \cong \alpha$}. x &\longmapsto & \text{the unique $\alpha \in \Ord$ such that $x \cong \alpha$}.
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\end{example} \end{example}
}{}
\begin{definition} \begin{definition}
A \vocab{prewellordering} $\preceq$ A \vocab{prewellordering} $\preceq$
@ -141,7 +144,7 @@ is a map $\phi\colon C \to \Ord$.
\item reflexive, \item reflexive,
\item transitive, \item transitive,
\item total (any two $x,y$ are comparable), \item total (any two $x,y$ are comparable),
\item $\prec$ ($x \prec y \iff x \preceq y \land y \not\preceq x$) is well-founded, \item $\prec$ \gist{($x \prec y \iff x \preceq y \land y \not\preceq x$)}{} is well-founded,
in the sense that there are no descending infinite chains. in the sense that there are no descending infinite chains.
\end{itemize} \end{itemize}
\end{definition} \end{definition}
@ -149,7 +152,7 @@ is a map $\phi\colon C \to \Ord$.
\begin{itemize} \begin{itemize}
\item A prewellordering may not be a linear order since \item A prewellordering may not be a linear order since
it is not necessarily antisymmetric. it is not necessarily antisymmetric.
\item The linearly ordered wellfounded sets are exactly the wellordered sets. %\item The linearly ordered wellfounded sets are exactly the wellordered sets.
\item Modding out $x \sim y :\iff x \preceq y \land y \preceq x$ \item Modding out $x \sim y :\iff x \preceq y \land y \preceq x$
turns a prewellordering into a wellordering. turns a prewellordering into a wellordering.
\end{itemize} \end{itemize}
@ -163,11 +166,11 @@ between downwards-closed ranks and prewellorderings:
\phi_{\preceq}&\longmapsfrom& \preceq, \phi_{\preceq}&\longmapsfrom& \preceq,
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
where $\phi_\preceq(x)$ is defined as where $\phi_\preceq(x)$ is defined as
\begin{IEEEeqnarray*}{rCl} \gist{\begin{IEEEeqnarray*}{rCl}
\phi_{\preceq}(x) &\coloneqq &0 \text{ if $x$ is minimal},\\ \phi_{\preceq}(x) &\coloneqq &0 \text{ if $x$ is minimal},\\
\phi_{\preceq}(x) &\coloneqq & \sup \{\phi_{\preceq}(y) + 1 : y \prec x\}, \phi_{\preceq}(x) &\coloneqq & \sup \{\phi_{\preceq}(y) + 1 : y \prec x\},
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
i.e. i.e.}{}
\[ \[
\phi_{\preceq}(x) = \otp\left(\faktor{\{y \in C : y \prec x\}}{\sim}\right). \phi_{\preceq}(x) = \otp\left(\faktor{\{y \in C : y \prec x\}}{\sim}\right).
\] \]

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inputs/lecture_14.tex
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@ -1,5 +1,6 @@
\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 $C$.