Josia Pietsch
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190 lines
7.1 KiB
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190 lines
7.1 KiB
TeX
\lecture{11}{2023-11-21}{}
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\begin{refproof}{thm:lusinsouslin}
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Note that $B_{(n_0,\ldots, n_k)} \subseteq B_{(n_0,\ldots, n_k)}^\ast \subseteq \overline{B_{n_0,\ldots, n_k}}$.
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We want to show that
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\[
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f(A) = \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast.
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\]
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Let $x \in f(A)$.
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Then take $a \in A$ such that $x = f(a)$.
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Then
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\[x \in \bigcap_k \underbrace{B_{a\defon{k}}}_{= f(A \cap N_{a\defon{k}})} \subseteq \bigcap_{k} B^\ast_{a\defon{k}}.\]
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This gives $f(A) \subseteq \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast$.
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If $x \in \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast$,
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Then there is a unique $a$ such that
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$x \in \bigcap_k B^\ast_{a\defon{k}}$.
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\begin{claim}
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$a \in A$.
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\end{claim}
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\begin{subproof}
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We have $B^\ast_{a\defon{k}} \subseteq \overline{B_{a\defon{k}}}$.
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So $x \in \bigcap_k \overline{B_{a\defon{k}}}$.
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In particular, $B_{a\defon{k}} \neq \emptyset$
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for all $k$.
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So for all $k$ we get that $A \cap N_{a\defon{k}} \neq \emptyset$.
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But $A$ is closed and $N_{a\defon{k}}$
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is clopen for all $k$.
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We have $\{a\} = \bigcup_k N_{a\defon{k}}$,
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so $a \in A$.
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\end{subproof}
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\begin{claim}
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$f(a) = x$.
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\end{claim}
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\begin{subproof}
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We have $f(a) \in \bigcap_k B_{a\defon{k}}$.
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Suppose $f(a) \neq x$.
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Pick $U \ni f(a)$ open
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such that $x \not\in \overline{U}$.
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By continuity of $f$,
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we get that $f(N_{a\defon{k_0}}) \subseteq U$
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for $k_0$ large enough.
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So $x \not\in \overline{f(N_{a\defon{k_0}})}$.
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In particular
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$x \not\in \overline{f(N_{a\defon{k_0}})} = \overline{B_{a\defon{k_0}}} \supset B^\ast_{a\defon{k_0}}$.
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But $x \in \bigcap_k B^\ast_{a\defon{k}} \lightning$.
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\end{subproof}
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\end{refproof}
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\begin{corollary}[of the \yaref{thm:lusinseparation}]
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\yalabel{Corollary of the Lusin Separation Theorem}{Lusin Separation}{cor:lusinseparation}
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Let $X$ be Polish.
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Let $A_1, A_2, A_3,\ldots \subseteq X$ be analytic
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and pairwise disjoint.
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Then there are pairwise disjoint Borel sets $B_i \supseteq A_i$.
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\end{corollary}
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\begin{proof}
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For all $i$, let $B_i, C_i$
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be disjoint Borel sets,
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such that $A_i \subseteq B_i$
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and $\bigcup_{j \neq i} A_j \subseteq C_i$.
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Take $D_i \coloneqq B_i \cap \bigcap_{j \neq i} C_j$.
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\end{proof}
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\begin{theorem}[\vocab{Borel Schröder-Bernstein}]
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\yalabel{Schröder-Bernstein for Borel sets}{Schröder-Bernstein}{thm:bsb}
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Let $A, B$ be Borel in some Polish spaces.
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Suppose that there are Borel embeddings
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$f\colon A \hookrightarrow B$
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and $g\colon B \hookrightarrow A$.
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Then $A$ and $B$ are Borel isomorphic.
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\end{theorem}
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\begin{proof}
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Cf.~\yaref{s7e4}.
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\end{proof}
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\begin{theorem}[\vocab{Isomorphism Theorem}]
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\yalabel{Isomorphism Theorem}{Isomorphism Thm.}{thm:isomorphism}
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Let $X, Y$ be Borel in some Polish spaces.
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Then $X$ is Borel isomorphic
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to $Y$ iff $|X| = |Y|$.
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\end{theorem}
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\begin{proof}
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$\implies$ is clear.
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Suppose that $|X| = |Y| \le \aleph_0$,
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then any bijection suffices,
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since all subsets are Borel.
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If $|X| = |Y| > \aleph_0$,
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then they must have cardinality $\fc$,
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since we can embed the Cantor space.
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It suffices to show that if $X$ is an uncountable
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Polish space and $\cC = 2^\omega$ the Cantor space,
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then they are Borel isomorphic.
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There is $2^\omega \hookrightarrow X$ Borel
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(continuous wrt.~the topology of $X$)
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On the other hand
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\[
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X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC
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\]
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\footnotetext{cf.~\yaref{s2e4}}
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For the first inclusion,
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recall that there is a continuous bijection $b\colon D \to X$,
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where $D \overset{\text{closed}}{\subseteq} \cN$.
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Consider $b^{-1}$.
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Whenever $B \subseteq X$ is Borel,
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we have that $b^{-1}(B)$ is Borel,
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since $b$ is continuous.
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For $A \subseteq D$ Borel
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be get by \yaref{thm:lusinsouslin},
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that $b$ with respect to $b(A)$
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is Borel,
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since $b\defon{A}$ is injective.
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Hence \yaref{thm:bsb} can be applied.
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\end{proof}
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\subsection{The Projective Hierarchy}
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% https://q.uiver.app/#q=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\[\begin{tikzcd}
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& {\Sigma^1_1(X)} && {\Sigma^1_2(X)} \\
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{\Delta^1_1(X)} && {\Delta^1_2(X)} \\
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& {\Pi^1_1(X)} && {\Pi^1_2(X)}
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\arrow["\subseteq", hook, from=2-1, to=1-2]
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\arrow["\subseteq"', hook, from=2-1, to=3-2]
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\arrow["\subseteq"', hook, from=3-2, to=2-3]
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\arrow["\subseteq", hook, from=1-2, to=2-3]
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\arrow["\subseteq", hook, from=2-3, to=1-4]
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\arrow["\subseteq", hook, from=2-3, to=3-4]
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\end{tikzcd}\]
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\begin{definition}
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Let $X$ be a Polish space.
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We define
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\begin{IEEEeqnarray*}{rCl}
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\Delta^1_n(X) &\coloneqq& \Sigma^1_n(X) \cap \Pi^1_n(X)\\
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\Pi^1_n(X) &=& \{A \subseteq X : X \setminus A \in \Sigma^1_n(X)\}\\
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\Sigma^1_{n+1}(X) &=& \{ A \subseteq X : \exists B \in \Pi^1_n(X \times \cN) .~A = \proj_X[B]\}
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\end{IEEEeqnarray*}
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\end{definition}
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\begin{theorem}
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Every analytic and every coanalytic set
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has the Baire property.
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\end{theorem}
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We will not proof this in this lecture.
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\subsection{Ill-Founded Trees}
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\gist{%
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Recall that a \vocab{tree} on $\N$ is a subset of
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$\N^{<\N}$
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closed under taking initial segments.
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We now identify trees with their characteristic functions,
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i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}%
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{We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:}
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\begin{IEEEeqnarray*}{rCl}
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\One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\
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x &\longmapsto & \begin{cases}
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1 &: x \in T,\\
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0 &: x \not\in T.
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\end{cases}
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\end{IEEEeqnarray*}
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\gist{Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.}{}
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Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
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\begin{observe}
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$\Tr \subseteq {2^{\N}}^{<\N}$ is closed
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(where we take the topology of the Cantor space).
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\end{observe}
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\gist{%
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Indeed, for any $ s \in \N^{<\N}$
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we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
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and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
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Boolean combinations of such sets are clopen as well.
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In particular for $s$ fixed,
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we have that
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\[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
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is clopen in ${2^{\N}}^{<\N}$.
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}{}
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