From 820f756bd4fd9d9cdc5b954a2f2da06ac0964450 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 17 Nov 2023 15:00:39 +0100 Subject: [PATCH] lecture 10 --- inputs/lecture_10.tex | 178 ++++++++++++++++++++++++++++++++++++++++++ logic3.tex | 1 + 2 files changed, 179 insertions(+) create mode 100644 inputs/lecture_10.tex diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex new file mode 100644 index 0000000..a01ab2b --- /dev/null +++ b/inputs/lecture_10.tex @@ -0,0 +1,178 @@ +\lecture{10}{2023-11-17}{} +\todo{Start a new subsection here?} +\begin{theorem}[\vocab{Lusin separation theorem}] + \yalabel{Lusin Separation Theorem}{Lusin Separation Thm.}{thm:lusinseparation} + Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic. + Then there is a Borel set $C$, + such that $A \subseteq C$ and $C \cap B = \emptyset$. +\end{theorem} +\begin{corollary} + Let $X$ be Polish. + Then + \[\cB(X) = \Delta^1_1(X),\] + where $\Delta^1_1(X) \coloneqq \Sigma^1_1(X) \cap \Pi_1^1(X)$. +\end{corollary} +\begin{proof} + Clearly $\cB(X) \subseteq \Delta^1_1(X)$. + + Let $A \in \Delta^1_1(X)$. + Then $A, X \setminus A \in \Sigma^1_1(X)$. + These can be separated by a Borel set $C$, + but then $A = C$, hence $A \in \cB(X)$. +\end{proof} + +For the proof of the \yaref{thm:lusinseparation}, +we need the following definition: +\begin{definition} + Let $X$ be Polish, + $P, Q \subseteq X$. + We say that $P, Q$ are \vocab{Borel-separable}, + if there exists $R \in \cB(X)$, + such that $P \subseteq R, Q \cap R = \emptyset$. +\end{definition} +\begin{lemma} + \label{lem:lusinsephelp} + If $P = \bigcup_{m < \omega} P_m$, $Q = \bigcup_{n < \omega} Q_n$ are such that + for any $m, n$ the sets $P_m$ and $Q_n$ are Borel separable. +\end{lemma} +\begin{proof} + For all $m, n$ pick $R_{m,n}$ Borel, + such that $P_m \subseteq R_{m,n}$ + and $Q_n \cap R_{m,n} = \emptyset$. + Then $R = \bigcup_m \bigcap_n R_{m,n}$ + has the desired property + that $R \subseteq R$ and $R \cap Q = \emptyset$. +\end{proof} + +\begin{notation} + For $s \in \omega^{<\omega}$ + be write $\cN_s \coloneqq \{x \in \cN : x \supseteq s\}$. +\end{notation} + +\begin{refproof}{thm:lusinseparation} + Let $X$ be Polish, + and $A, B \subseteq X$ analytic + such that $A \cap B = \emptyset$ + Then there are continuous surjections + $f\colon \cN \twoheadrightarrow A \subseteq X$ + and $g\colon \twoheadrightarrow B \subseteq X$. + + Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$. + Note that $A_s = \bigcup_m A_{s\concat m}$ + and $B_ns = \bigcup_{n < \omega} B_{s\concat n}$. + + In particular + $A = \bigcup_{m < \omega} A_{\underbrace{\langle m \rangle}_{\in \omega^1}}$ + and $B = \bigcup_{n < \omega} B_{\langle n \rangle}$. + Towards a contradiction suppose that + $A$ and $B$ are not Borel separable. + Then by \yaref{lem:lusinsephelp}, + there exist $m,n$ such that $A_{\langle m \rangle}$ and $B_{\langle n \rangle}$ + can't be separated. + Since $A_{\langle m \rangle} = \bigcup_i A_{\langle m, i \rangle}$ + and similarly for $B$, + there exist $i,j$ such that + $A_{\langle m,i \rangle}$ and $B_{\langle n, j\rangle}$ + are not Borel separable. + + Recursively, we find sequences $x,y \in \cN$, + such that $A_{x\defon{n}}$ and $B_{y\defon{n}}$ + are not Borel separable for any $n < \omega$. + So $f(x) \in A$ and $g(y) \in B$. + + Recall that $A_{x\defon{n}} = f(\cN_{x\defon{n}}$ + and $B_{y\defon{n}} = g(\cN_{x\defon{n}})$. + + Since $A \cap B = \emptyset$, + we get that $f(x) \neq g(y)$. + Let $U,V$ be disjoint open such that + $f(x) \in U, g(y) \in V$. + As $f$ and $g$ are continuous, + $U \subseteq f(\cN_{x\defon{n}})$ + and + $V \subseteq g(\cN_{x\defon{n}})$ + for $n $ large enough. + Then $U$ separates $A_{x\defon{n_0}}$ + and $V$ separates $B_{y\defon{n_0}}$, + contradicting the choice of $x$ and $y$. +\end{refproof} +\begin{theorem}[Lusin-Souslin] + \yalabel{Lusin-Souslin}{Lusin-Souslin}{thm:lusinsouslin} + Let $X, Y$ be Polish + and $f\colon X \to Y$ Borel. + Let $A \in \cB(X)$ + such that $f\defon{A}$ is injective. + Then $f(A)$ is Borel. +\end{theorem} +\begin{proof} + 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.} + and $X = \cN$ by \yaref{thm:bairetopolish}: + + +% https://q.uiver.app/#q=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 +\[\begin{tikzcd} + \cN \\ + Z & X & Y \\ + {h^{-1}(A)} & A & {f(A)} + \arrow["f", from=2-2, to=2-3] + \arrow["{f\defon{A}}"', hook, from=3-2, to=3-3] + \arrow[from=2-1, to=1-1] + \arrow["{h }", tail reversed, from=2-1, to=2-2] + \arrow["\subseteq"{description}, hook, from=3-1, to=2-1] + \arrow["\subseteq"{description}, hook, from=3-2, to=2-2] + \arrow["\subseteq"{description}, hook, from=3-3, to=2-3] + \arrow["{h\defon{A}}"', tail reversed, from=3-1, to=3-2] +\end{tikzcd}\] + + For $s \in \omega^{<\omega}$ write $B_s \coloneqq f(\cN_s \cap A)$. + + As in the previous proof we have + $B_\emptyset = f(A)$ + and $B_s = \bigcup_{n < \omega} B_{s\concat n}$ + for every $s \in \omega^{<\omega}$. + + Note that + \begin{itemize} + \item $\forall n.~\forall s.~ B_{s\concat n } \subseteq B_s$ and + \item $\forall n \neq n'.~\forall s.~B_{s\concat n} \cap B_{s \concat n'} = \emptyset$. + \end{itemize} + The second point follows from injectivity of $f$ + and the fact that $\cN_{s \concat n} \cap \cN_{s\concat n'} = \emptyset$. + In particular, the $(B_s)$ for a Lusin scheme. + + Note that $f(A) = \bigcup_{s \in \omega^k} B_s$ + for every $k <\omega$, + thus $f(A) = \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s$. + We want to find $B_s^\ast \in \cB(X)$ for $s \in \omega^{<\omega}$, + such that the $B_s^\ast$ form a Lusin scheme + and still + \[ + f(A) = \bigcap_{k < \omega} \bigcup_{s \in \omega^{<\omega}} B_s^\ast. + \] + The existence of such $B_s^\ast$ implies that + $f(A)$ is Borel. + + By a generalization of \yaref{thm:lusinseparation},\todo{TODO} + for all $k < \omega$, + we can separate the collection of disjoint analytic sets $\{B_s : s \in \omega^k\}$ + Borel sets, + i.e. there are disjoint Borel sets $(C_s)_{s \in \omega^k}$ + such that $B_s \subseteq C_s$. + + Using this, we get a Lusin scheme $(B'_s)_{s \in \omega^{<\omega}}$ + such that the $B_s'$ are Borel, + $B'_{\emptyset} = Y$ and $B_s \subseteq B_s'$: + Set $B'_\emptyset = Y$ and $B'_{s\concat n } = B'_s \cap C_{s \concat n}$. + However the $B'_s$ might be to large. + + We define another Lusin scheme $(B^\ast_s)_s$ as follows: + Let $B^\ast_\emptyset \coloneqq Y$, + and for $s \in \omega^{<\omega}$, $n < \omega$ + \[ + B^\ast_{s \concat n} = B'_{s \concat n} \cap \overline{B_{s \concat n}} \cap B^\ast_{s}. + \] + + + +\end{proof} diff --git a/logic3.tex b/logic3.tex index cecc7a0..cf0d291 100644 --- a/logic3.tex +++ b/logic3.tex @@ -33,6 +33,7 @@ \input{inputs/lecture_07} \input{inputs/lecture_08} \input{inputs/lecture_09} +\input{inputs/lecture_10}