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\subsection{The Lusin Separation Theorem}
\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 $C \cap B = \emptyset$.
Let $X$ be Polish.
\[\cB(X) = \Delta^1_1(X),\]
where $\Delta^1_1(X) \coloneqq \Sigma^1_1(X) \cap \Pi_1^1(X)$.
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 $A \in \cB(X)$.
For the proof of the \yaref{thm:lusinseparation},
we need the following 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 $P \subseteq R, Q \cap R = \emptyset$.
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,
then $P$ and $Q$ are Borel separable.
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 $P \subseteq R$ and $R \cap Q = \emptyset$.
For $s \in \omega^{<\omega}$
be write $\cN_s \coloneqq \{x \in \cN : x \supseteq s\}$.
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 $B_s = \bigcup_{n < \omega} B_{s\concat n}$.
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,
$U \subseteq f(\cN_{x\defon{n}})$
$V \subseteq g(\cN_{x\defon{n}})$
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 $y$.
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.
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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
\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]
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
\item $\forall n.~\forall s.~ B_{s\concat n } \subseteq B_s$ and
\item $\forall n \neq n'.~\forall s.~B_{s\concat n} \cap B_{s \concat n'} = \emptyset$.
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}.