\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} Cf.~\yaref{s7e4}. \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.~the topology of $X$) On the other hand \[ X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC \] \footnotetext{cf.~\yaref{s2e4}} 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 $A \subseteq D$ Borel be get by \yaref{thm:lusinsouslin}, that $b$ with respect to $b(A)$ is Borel, since $b\defon{A}$ is injective. Hence \yaref{thm:bsb} can be applied. \end{proof} \subsection{The 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{Ill-Founded Trees} \gist{% 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 $T \subseteq \N^{<\N}$}% {We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:} \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*} \gist{Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.}{} Let \vocab{$\Tr$} $ \coloneqq \{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} \gist{% 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 ${2^{\N}}^{<\N}$. }{}