From eb87abe4725aa223401ce07559af8d50bd35e9e1 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 21 Nov 2023 12:34:38 +0100 Subject: [PATCH] lecture 11 --- inputs/lecture_10.tex | 9 +- inputs/lecture_11.tex | 190 +++++++++++++++++++++++++++++++++++++++++ inputs/tutorial_05.tex | 23 +++++ logic.sty | 2 + logic3.tex | 1 + 5 files changed, 221 insertions(+), 4 deletions(-) create mode 100644 inputs/lecture_11.tex create mode 100644 inputs/tutorial_05.tex diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex index a01ab2b..d6191dc 100644 --- a/inputs/lecture_10.tex +++ b/inputs/lecture_10.tex @@ -1,7 +1,7 @@ \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} + \yalabel{Lusin Separation Theorem}{Lusin Separation}{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 $C \cap B = \emptyset$. @@ -104,7 +104,7 @@ we need the following definition: such that $f\defon{A}$ is injective. Then $f(A)$ is Borel. \end{theorem} -\begin{proof} +\begin{refproof}{thm:lusinsouslin} 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}: @@ -153,7 +153,7 @@ we need the following definition: The existence of such $B_s^\ast$ implies that $f(A)$ is Borel. - By a generalization of \yaref{thm:lusinseparation},\todo{TODO} + By the \yaref{cor:lusinseparation}, for all $k < \omega$, we can separate the collection of disjoint analytic sets $\{B_s : s \in \omega^k\}$ Borel sets, @@ -174,5 +174,6 @@ we need the following definition: \] + \phantom\qedhere -\end{proof} +\end{refproof} diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex new file mode 100644 index 0000000..f860080 --- /dev/null +++ b/inputs/lecture_11.tex @@ -0,0 +1,190 @@ +\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 ${2^{\N}}^{<\N}$. + + + diff --git a/inputs/tutorial_05.tex b/inputs/tutorial_05.tex new file mode 100644 index 0000000..6ed47ee --- /dev/null +++ b/inputs/tutorial_05.tex @@ -0,0 +1,23 @@ +\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} + + diff --git a/logic.sty b/logic.sty index 5fbc0b7..d708fe0 100644 --- a/logic.sty +++ b/logic.sty @@ -126,6 +126,8 @@ \DeclareSimpleMathOperator{trcl} \DeclareSimpleMathOperator{tcl} +\DeclareSimpleMathOperator{Tr} + \newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}} %https://tex.stackexchange.com/questions/73437/how-do-i-typeset-the-concatenation-of-strings-properly diff --git a/logic3.tex b/logic3.tex index cf0d291..1615c2f 100644 --- a/logic3.tex +++ b/logic3.tex @@ -34,6 +34,7 @@ \input{inputs/lecture_08} \input{inputs/lecture_09} \input{inputs/lecture_10} +\input{inputs/lecture_11}