w23-logic-3/inputs/lecture_10.tex

\subsection{The Lusin Separation Theorem}

\lecture{10}{2023-11-17}{}
\begin{theorem}[\vocab{Lusin separation theorem}]
    \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 $C \cap B = \emptyset$.
\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 $g\colon \cN \twoheadrightarrow B \subseteq X$.

    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{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}:


% 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 $(B_s)$ form 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 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,
     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 $s \in \omega^{<\omega}$, $n < \omega$
     \[
     B^\ast_{s \concat n} = B'_{s \concat n} \cap \overline{B_{s \concat n}} \cap B^\ast_{s}.
     \]


     \phantom\qedhere

\end{refproof}