180 lines
6.9 KiB
TeX
180 lines
6.9 KiB
TeX
\subsection{The Lusin Separation Theorem}
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\lecture{10}{2023-11-17}{}
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\begin{theorem}[\vocab{Lusin separation theorem}]
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\yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
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Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic.
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Then there is a Borel set $C$,
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such that $A \subseteq C$ and $C \cap B = \emptyset$.
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\end{theorem}
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\begin{corollary}
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Let $X$ be Polish.
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Then
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\[\cB(X) = \Delta^1_1(X),\]
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where $\Delta^1_1(X) \coloneqq \Sigma^1_1(X) \cap \Pi_1^1(X)$.
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\end{corollary}
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\begin{proof}
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Clearly $\cB(X) \subseteq \Delta^1_1(X)$.
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Let $A \in \Delta^1_1(X)$.
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Then $A, X \setminus A \in \Sigma^1_1(X)$.
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These can be separated by a Borel set $C$,
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but then $A = C$, hence $A \in \cB(X)$.
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\end{proof}
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For the proof of the \yaref{thm:lusinseparation},
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we need the following definition:
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\begin{definition}
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Let $X$ be Polish,
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$P, Q \subseteq X$.
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We say that $P, Q$ are \vocab{Borel-separable},
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if there exists $R \in \cB(X)$,
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such that $P \subseteq R, Q \cap R = \emptyset$.
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\end{definition}
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\begin{lemma}
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\label{lem:lusinsephelp}
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If $P = \bigcup_{m < \omega} P_m$, $Q = \bigcup_{n < \omega} Q_n$ are such that
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for any $m, n$ the sets $P_m$ and $Q_n$ are Borel separable.
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\end{lemma}
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\begin{proof}
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For all $m, n$ pick $R_{m,n}$ Borel,
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such that $P_m \subseteq R_{m,n}$
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and $Q_n \cap R_{m,n} = \emptyset$.
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Then $R = \bigcup_m \bigcap_n R_{m,n}$
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has the desired property
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that $R \subseteq R$ and $R \cap Q = \emptyset$.
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\end{proof}
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\begin{notation}
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For $s \in \omega^{<\omega}$
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be write $\cN_s \coloneqq \{x \in \cN : x \supseteq s\}$.
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\end{notation}
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\begin{refproof}{thm:lusinseparation}
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Let $X$ be Polish,
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and $A, B \subseteq X$ analytic
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such that $A \cap B = \emptyset$
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Then there are continuous surjections
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$f\colon \cN \twoheadrightarrow A \subseteq X$
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and $g\colon \cN \twoheadrightarrow B \subseteq X$.
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Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$.
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Note that $A_s = \bigcup_m A_{s\concat m}$
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and $B_ns = \bigcup_{n < \omega} B_{s\concat n}$.
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In particular
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$A = \bigcup_{m < \omega} A_{\underbrace{\langle m \rangle}_{\in \omega^1}}$
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and $B = \bigcup_{n < \omega} B_{\langle n \rangle}$.
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Towards a contradiction suppose that
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$A$ and $B$ are not Borel separable.
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Then by \yaref{lem:lusinsephelp},
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there exist $m,n$ such that $A_{\langle m \rangle}$ and $B_{\langle n \rangle}$
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can't be separated.
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Since $A_{\langle m \rangle} = \bigcup_i A_{\langle m, i \rangle}$
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and similarly for $B$,
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there exist $i,j$ such that
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$A_{\langle m,i \rangle}$ and $B_{\langle n, j\rangle}$
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are not Borel separable.
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Recursively, we find sequences $x,y \in \cN$,
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such that $A_{x\defon{n}}$ and $B_{y\defon{n}}$
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are not Borel separable for any $n < \omega$.
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So $f(x) \in A$ and $g(y) \in B$.
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Recall that $A_{x\defon{n}} = f(\cN_{x\defon{n}}$
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and $B_{y\defon{n}} = g(\cN_{x\defon{n}})$.
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Since $A \cap B = \emptyset$,
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we get that $f(x) \neq g(y)$.
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Let $U,V$ be disjoint open such that
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$f(x) \in U, g(y) \in V$.
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As $f$ and $g$ are continuous,
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$U \subseteq f(\cN_{x\defon{n}})$
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and
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$V \subseteq g(\cN_{x\defon{n}})$
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for $n $ large enough.
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Then $U$ separates $A_{x\defon{n_0}}$
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and $V$ separates $B_{y\defon{n_0}}$,
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contradicting the choice of $x$ and $y$.
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\end{refproof}
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\begin{theorem}[Lusin-Souslin]
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\yalabel{Lusin-Souslin}{Lusin-Souslin}{thm:lusinsouslin}
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Let $X, Y$ be Polish
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and $f\colon X \to Y$ Borel.
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Let $A \in \cB(X)$
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such that $f\defon{A}$ is injective.
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Then $f(A)$ is Borel.
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\end{theorem}
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\begin{refproof}{thm:lusinsouslin}
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W.l.o.g.~suppose that $f$ is continuous,
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$A$ is closed\footnote{We might even assume that $A$ is clopen, but we only need closed.}
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and $X = \cN$ by \yaref{thm:bairetopolish}:
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% https://q.uiver.app/#q=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\[\begin{tikzcd}
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\cN \\
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Z & X & Y \\
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{h^{-1}(A)} & A & {f(A)}
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\arrow["f", from=2-2, to=2-3]
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\arrow["{f\defon{A}}"', hook, from=3-2, to=3-3]
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\arrow[from=2-1, to=1-1]
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\arrow["{h }", tail reversed, from=2-1, to=2-2]
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\arrow["\subseteq"{description}, hook, from=3-1, to=2-1]
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\arrow["\subseteq"{description}, hook, from=3-2, to=2-2]
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\arrow["\subseteq"{description}, hook, from=3-3, to=2-3]
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\arrow["{h\defon{A}}"', tail reversed, from=3-1, to=3-2]
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\end{tikzcd}\]
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For $s \in \omega^{<\omega}$ write $B_s \coloneqq f(\cN_s \cap A)$.
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As in the previous proof we have
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$B_\emptyset = f(A)$
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and $B_s = \bigcup_{n < \omega} B_{s\concat n}$
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for every $s \in \omega^{<\omega}$.
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Note that
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\begin{itemize}
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\item $\forall n.~\forall s.~ B_{s\concat n } \subseteq B_s$ and
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\item $\forall n \neq n'.~\forall s.~B_{s\concat n} \cap B_{s \concat n'} = \emptyset$.
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\end{itemize}
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The second point follows from injectivity of $f$
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and the fact that $\cN_{s \concat n} \cap \cN_{s\concat n'} = \emptyset$.
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In particular, the $(B_s)$ form a Lusin scheme.
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Note that $f(A) = \bigcup_{s \in \omega^k} B_s$
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for every $k <\omega$,
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thus $f(A) = \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s$.
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We want to find $B_s^\ast \in \cB(X)$ for $s \in \omega^{<\omega}$,
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such that the $B_s^\ast$ form a Lusin scheme
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and still
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\[
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f(A) = \bigcap_{k < \omega} \bigcup_{s \in \omega^{<\omega}} B_s^\ast.
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\]
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The existence of such $B_s^\ast$ implies that
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$f(A)$ is Borel.
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By the \yaref{cor:lusinseparation},
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for all $k < \omega$,
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we can separate the collection of disjoint analytic sets $\{B_s : s \in \omega^k\}$
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Borel sets,
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i.e. there are disjoint Borel sets $(C_s)_{s \in \omega^k}$
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such that $B_s \subseteq C_s$.
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Using this, we get a Lusin scheme $(B'_s)_{s \in \omega^{<\omega}}$
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such that the $B_s'$ are Borel,
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$B'_{\emptyset} = Y$ and $B_s \subseteq B_s'$:
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Set $B'_\emptyset = Y$ and $B'_{s\concat n } = B'_s \cap C_{s \concat n}$.
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However the $B'_s$ might be to large.
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We define another Lusin scheme $(B^\ast_s)_s$ as follows:
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Let $B^\ast_\emptyset \coloneqq Y$,
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and for $s \in \omega^{<\omega}$, $n < \omega$
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\[
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B^\ast_{s \concat n} = B'_{s \concat n} \cap \overline{B_{s \concat n}} \cap B^\ast_{s}.
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\]
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\phantom\qedhere
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\end{refproof}
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