w23-logic-3/inputs/lecture_11.tex

\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.~to the topology of $X$)
    On the other hand
    \[
    X \hookrightarrow\cN \overset{\text{continuous embedding}}{\hookrightarrow}\cC
    \]
    \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{The Projective Hierarchy}

        
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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}


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