lecture 11
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\lecture{10}{2023-11-17}{} \lecture{10}{2023-11-17}{}
\todo{Start a new subsection here?} \todo{Start a new subsection here?}
\begin{theorem}[\vocab{Lusin separation theorem}] \begin{theorem}[\vocab{Lusin separation theorem}]
\yalabel{Lusin Separation Theorem}{Lusin Separation Thm.}{thm:lusinseparation} \yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic. Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic.
Then there is a Borel set $C$, Then there is a Borel set $C$,
such that $A \subseteq C$ and $C \cap B = \emptyset$. such that $A \subseteq C$ and $C \cap B = \emptyset$.
@ -104,7 +104,7 @@ we need the following definition:
such that $f\defon{A}$ is injective. such that $f\defon{A}$ is injective.
Then $f(A)$ is Borel. Then $f(A)$ is Borel.
\end{theorem} \end{theorem}
\begin{proof} \begin{refproof}{thm:lusinsouslin}
W.l.o.g.~suppose that $f$ is continuous W.l.o.g.~suppose that $f$ is continuous
$A$ is closed,\footnote{We might even assume that $A$ is clopen, but we only need closed.} $A$ is closed,\footnote{We might even assume that $A$ is clopen, but we only need closed.}
and $X = \cN$ by \yaref{thm:bairetopolish}: and $X = \cN$ by \yaref{thm:bairetopolish}:
@ -153,7 +153,7 @@ we need the following definition:
The existence of such $B_s^\ast$ implies that The existence of such $B_s^\ast$ implies that
$f(A)$ is Borel. $f(A)$ is Borel.
By a generalization of \yaref{thm:lusinseparation},\todo{TODO} By the \yaref{cor:lusinseparation},
for all $k < \omega$, for all $k < \omega$,
we can separate the collection of disjoint analytic sets $\{B_s : s \in \omega^k\}$ we can separate the collection of disjoint analytic sets $\{B_s : s \in \omega^k\}$
Borel sets, Borel sets,
@ -174,5 +174,6 @@ we need the following definition:
\] \]
\phantom\qedhere
\end{proof} \end{refproof}

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\lecture{11}{2023-11-21}{}
\begin{refproof}{thm:lusinsouslin}
Note that $B_{(n_0,\ldots, n_k)} \subseteq B_{(n_0,\ldots, n_k)}^\ast \subseteq \overline{B_{n_0,\ldots, n_k}}$.
We want to show that
\[
f(A) = \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast.
\]
Let $x \in f(A)$.
Then take $a \in A$ such that $x = f(a)$.
Then
\[x \in \bigcap_k \underbrace{B_{a\defon{k}}}_{= f(A \cap N_{a\defon{k}})} \subseteq \bigcap_{k} B^\ast_{a\defon{k}}.\]
This gives $f(A) \subseteq \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast$.
If $x \in \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast$,
Then there is a unique $a$ such that
$x \in \bigcap_k B^\ast_{a\defon{k}}$.
\begin{claim}
$a \in A$.
\end{claim}
\begin{subproof}
We have $B^\ast_{a\defon{k}} \subseteq \overline{B_{a\defon{k}}}$.
So $x \in \bigcap_k \overline{B_{a\defon{k}}}$.
In particular, $B_{a\defon{k}} \neq \emptyset$
for all $k$.
So for all $k$ we get that $A \cap N_{a\defon{k}} \neq \emptyset$.
But $A$ is closed and $N_{a\defon{k}}$
is clopen for all $k$.
We have $\{a\} = \bigcup_k N_{a\defon{k}}$,
so $a \in A$.
\end{subproof}
\begin{claim}
$f(a) = x$.
\end{claim}
\begin{subproof}
We have $f(a) \in \bigcap_k B_{a\defon{k}}$.
Suppose $f(a) \neq x$.
Pick $U \ni f(a)$ open
such that $x \not\in \overline{U}$.
By continuity of $f$,
we get that $f(N_{a\defon{k_0}}) \subseteq U$
for $k_0$ large enough.
So $x \not\in \overline{f(N_{a\defon{k_0}})}$.
In particular
$x \not\in \overline{f(N_{a\defon{k_0}})} = \overline{B_{a\defon{k_0}}} \supset B^\ast_{a\defon{k_0}}$.
But $x \in \bigcap_k B^\ast_{a\defon{k}} \lightning$.
\end{subproof}
\end{refproof}
\begin{corollary}[of the \yaref{thm:lusinseparation}]
\yalabel{Corollary of the Lusin Separation Theorem}{Lusin Separation}{cor:lusinseparation}
Let $X$ be Polish.
Let $A_1, A_2, A_3,\ldots \subseteq X$ be analytic
and pairwise disjoint.
Then there are pairwise disjoint Borel sets $B_i \supseteq A_i$.
\end{corollary}
\begin{proof}
For all $i$, let $B_i, C_i$
be disjoint Borel sets,
such that $A_i \subseteq B_i$
and $\bigcup_{j \neq i} A_j \subseteq C_i$.
Take $D_i \coloneqq B_i \cap \bigcap_{j \neq i} C_j$.
\end{proof}
\begin{theorem}[\vocab{Borel Schröder-Bernstein}]
\yalabel{Schröder-Bernstein for Borel sets}{Schröder-Bernstein}{thm:bsb}
Let $A, B$ be Borel in some Polish spaces.
Suppose that there are Borel embeddings
$f\colon A \hookrightarrow B$
and $g\colon B \hookrightarrow A$.
Then $A$ and $B$ are Borel isomorphic.
\end{theorem}
\begin{proof}
\todo{Homework}
\end{proof}
\begin{theorem}[\vocab{Isomorphism Theorem}]
\yalabel{Isomorphism Theorem}{Isomorphism Thm.}{thm:isomorphism}
Let $X, Y$ be Borel in some Polish spaces.
Then $X$ is Borel isomorphic
to $Y$ iff $|X| = |Y|$.
\end{theorem}
\begin{proof}
$\implies$ is clear.
Suppose that $|X| = |Y| \le \aleph_0$,
then any bijection suffices,
since all subsets are Borel.
If $|X| = |Y| > \aleph_0$,
then they must have cardinality $\fc$,
since we can embed the Cantor space.
It suffices to show that if $X$ is an uncountable
Polish space and $\cC = 2^\omega$ the Cantor space,
then they are Borel isomorphic.
There is $2^\omega \hookrightarrow X$ Borel
(continuous wrt.~to the topology of $X$)
On the other hand
\[
X \hookrightarrow\cN \hookrightarrow[\text{continuous embedding}]\cC
\]
\todo{second inclusion was on a homework sheet}
For the first inclusion,
recall that there is a continuous bijection $b\colon D \to X$,
where $D \overset{\text{closed}}{\subseteq} \cN$.
Consider $b^{-1}$.
Whenever $B \subseteq X$ is Borel,
we have that $b^{-1}(B)$ is Borel,
since $b$ is continuous.
For $A \subseteq \cN$ is Borel,
we have that $b$ with respect to $b(A)$
is Borel,
since $b\defon{A}$ is injective,
by \yaref{thm:lusinsouslin}.
Hence \yaref{thm:bsb} can be applied.
\end{proof}
\subsection{Projective Hierarchy}
% https://q.uiver.app/#q=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
\[\begin{tikzcd}
& {\Sigma^1_1(X)} && {\Sigma^1_2(X)} \\
{\Delta^1_1(X)} && {\Delta^1_2(X)} \\
& {\Pi^1_1(X)} && {\Pi^1_2(X)}
\arrow["\subseteq", hook, from=2-1, to=1-2]
\arrow["\subseteq"', hook, from=2-1, to=3-2]
\arrow["\subseteq"', hook, from=3-2, to=2-3]
\arrow["\subseteq", hook, from=1-2, to=2-3]
\arrow["\subseteq", hook, from=2-3, to=1-4]
\arrow["\subseteq", hook, from=2-3, to=3-4]
\end{tikzcd}\]
\begin{definition}
Let $X$ be a Polish space.
We define
\begin{IEEEeqnarray*}{rCl}
\Delta^1_n(X) &\coloneqq& \Sigma^1_n(X) \cap \Pi^1_n(X)\\
\Pi^1_n(X) &=& \{A \subseteq X : X \setminus A \in \Sigma^1_n(X)\}\\
\Sigma^1_{n+1}(X) &=& \{ A \subseteq X : \exists B \in \Pi^1_n(X \times \cN) .~A = \proj_X[B]\}
\end{IEEEeqnarray*}
\end{definition}
\begin{theorem}
Every analytic and every coanalytic set
has the Baire property.
\end{theorem}
We will not proof this in this lecture.
\subsection{Trees} % TODO section?
Recall that a \vocab{tree} on $\N$ is a subset of
$\N^{<\N}$
closed under taking initial segments.
We now identify trees with their characteristic functions,
i.e.~we want to associate a tree $T \subseteq \N^{<\N}$
\begin{IEEEeqnarray*}{rCl}
\One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\
x &\longmapsto & \begin{cases}
1 &: x \in T,\\
0 &: x \not\in T.
\end{cases}
\end{IEEEeqnarray*}
Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.
Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
\begin{observe}
\[
\Tr \subseteq {2^{\N}}^{<\N}
\]
is closed (where we take the topology of the Cantor space).
\end{observe}
Indeed, for any $ s \in \N^{<\N}$
we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
Boolean combinations of such sets are clopen as well.
In particular for $s$ fixed,
we have that
\[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
is clopen in ${2^{\N}}^{<\N}$.

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\tutorial{05}{}{}
% Sheet 5 - 18.5 / 20
\subsection{Exercise 1}
Let $B \subseteq C$ be comeager.
Then $B = B_1 \cup B_2$,
where $B_1$ is dense $G_\delta$
and $B_2$ is meager.
\begin{fact}
$X$ is Baire iff every non-empty open set is non-meager.
In particular, let $X$ be Baire,
then $U \overset{\text{open}}{\subseteq} X$
is Baire.
\end{fact}
\subsection{Exercise 2}

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\DeclareSimpleMathOperator{trcl} \DeclareSimpleMathOperator{trcl}
\DeclareSimpleMathOperator{tcl} \DeclareSimpleMathOperator{tcl}
\DeclareSimpleMathOperator{Tr}
\newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}} \newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}
%https://tex.stackexchange.com/questions/73437/how-do-i-typeset-the-concatenation-of-strings-properly %https://tex.stackexchange.com/questions/73437/how-do-i-typeset-the-concatenation-of-strings-properly

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\input{inputs/lecture_08} \input{inputs/lecture_08}
\input{inputs/lecture_09} \input{inputs/lecture_09}
\input{inputs/lecture_10} \input{inputs/lecture_10}
\input{inputs/lecture_11}