lecture 11

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}

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 ${2^{\N}}^{<\N}$.
+
+
+

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}
+
+

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

logic3.tex

@@ -34,6 +34,7 @@
 \input{inputs/lecture_08}
 \input{inputs/lecture_09}
 \input{inputs/lecture_10}
+\input{inputs/lecture_11}