\subsection{Sheet 7}
\tutorial{08}{2023-12-05}{}
% 17 / 20
\nr 1
\begin{itemize}
    \item For $\xi = 1$ this holds by the definition of the
        subspace topology.

        We now use transfinite induction, to show that
        the statement holds for all $\xi$.
        Suppose that $\Sigma^0_{\zeta}(Y)$ and $\Pi^0_{\zeta}(Y)$
        are as claimed for all $\zeta < \xi$.

        Then
        \begin{IEEEeqnarray*}{rCl}
            \Sigma^0_\xi(Y) &=& \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(Y), \alpha_n < \xi\}\\
                            &=& \{\bigcup_{n < \omega} (A_n \cap Y) : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
                            &=& \{Y \cap \bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
                            &=& \{Y \cap A : A \in \Sigma^0_{\xi}(X)\}.
        \end{IEEEeqnarray*}
        and
        \begin{IEEEeqnarray*}{rCl}
            \Pi^0_\xi(Y) &=& \lnot \Sigma^0_\xi(Y)\\
                         &=& \{Y \setminus A : A \in \Sigma^0_\xi(Y)\}\\
                         &=& \{Y \setminus (A \cap Y) : A \in \Sigma^0_\xi(X)\}\\
                         &=& \{Y \cap (X \setminus A) : A \in \Sigma^0_\xi(X)\}\\
                         &=& \{Y \cap A : A \in \Pi^0_\xi(X)\}.
        \end{IEEEeqnarray*}
    \item Let $V \in \cB(Y)$.

        We show that $f^{-1}(V) \in \cB(Y)$,
        by induction on the minimal $\xi$ such that $V \in \Sigma_\xi^0$.
        For $\xi = 0$ this is clear.
        Suppose that we have already shown $f^{-1}(V') \in \cB(Y)$
        for all $V' \in \Sigma^0_\zeta$, $\zeta < \xi$.
        Then $f^{-1}(Y \setminus V')  = X \setminus f^{-1}(V') \in \cB(V)$,
        since complements of Borel sets are Borel.
        In particular, this also holds for $\Pi^0_\zeta$ sets
        and $\zeta < \xi$.
        Let $V \in \Sigma^0_\xi$.
        Then $V = \bigcap_{n} V_n$ for some $V_n \in \Pi^{0}_{\alpha_n}$,
        $\alpha_n < \xi$.
        In particular $f^{-1}(V) = \bigcup_n f^{-1}(V_n) \in \cB(X)$.
\end{itemize}

\nr 2


Recall \autoref{thm:analytic}.

Let $(A_i)_{i<\omega}$ be analytic subsets of a Polish space $X$.
Then there exists Polish spaces $Y_i$ and $f_i\colon Y_i \to X$
continuous such that $f_i(B_i) = A_i$
for some $B_i \in \cB(Y_i)$.

 \begin{itemize}
    \item $\bigcup_i A_i$ is analytic:
        Consider the Polish space $Y \coloneqq  \coprod_{i < \omega} Y_i$
        and $f \coloneqq \coprod_i f_i$, i.e.~
        $Y_i \ni y \mapsto f_i(y)$.
        $f$ is continuous,
        $\coprod_{i < \omega} B_i \in \cB(Y)$
        and
        \[f(\coprod_{i < \omega} B_i) = \bigcup_i A_i.\]
    \item $\bigcap_i A_i$ is analytic:
        % Let $Y_i$ be Polish such that $f_i(Y_i) = A_i$.
        % Let $Y \coloneqq \coprod Y_i$, $f = \coprod f_i$ and $Z = \prod Y_i$.
        % Note that $Y$ and $Z$ are Polish.
        % We can embed $Z$ into $Y^{\N}$.
        % 
        % Define a tree $T$ on $Y$ as follows:
        % $(y_0, \ldots, y_n) \in T$ iff
        % \begin{itemize}
        %     \item $\forall 0 \le i \le n.~ y_i \in Y_i$ and
        %     \item $\forall  i,j .~ f(y_i) = f(y_j)$.
        % \end{itemize}
        % 
        % Then $[T]$ consists of sequences $y = (y_n)$
        % such that $\forall j \in \N.~f(y) \in  \im (f_j)$,
        % so $\forall y \in [T].~f(y) \in \bigcap_{i \in \N} \im(f_i) = \bigcap_{i \in \N} A_i$.
        % $[T] \subseteq  i(Z) \subseteq Y^{\N}$,
        % and $[T]$ is closed.
        % 
        % 
        % Other solution:
        Let $Z = \prod Y_i$
        and let $D \subseteq  Z$
        be defined by
        \[
        D \coloneqq \{(y_n) : f_i(y_i) = f_j(y_j) ~ \forall i,j\}.
        \]
        $D$ is closed,
        at it is the preimage of the diagonal
        under $Z \xrightarrow{(f_0,f_1,\ldots)} X^{\N}$.
        Then $\bigcap A_i$ is the image of  $D$
        under $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
        
        \paragraph{Other solution}

        Let $F_n \subseteq X \times  \cN$ be closed,
        and $C \subseteq  X \times \cN^{\N}$ defined by
        \[
        C \coloneqq  \{(x,(y^{(n)}) ) : \forall n.~(x, y^{(n)}) \in  F_n\}.
        \]
        $C$ is closed
        and $\bigcap A_i = \proj_X(C)$.
\end{itemize}

\nr 3
\begin{lemma}
    Let $X$ be a second-countable topological space.
    Then every base of $X$ contains a countable subset which
    is also a base of $X$.
\end{lemma}
\begin{proof}
    Let $\cC = \{C_n : n < \omega\}$
    be a countable base of $X$
    and let $\cB = \{B_i: i \in I\}$
    be a base of $X$ with (possibly uncountable) index set $I$.

    Fix $n < \omega$. It suffices to show that $C_n$ is a union
    of countably many elements of $\cB$.
    As  $\cB$ is a base, $C_n = \bigcup_{j \in J} B_j$
    for some $J \subseteq I$.
    Since $\cC$ is a base,
    there exists $M_j \subseteq \N$ such that $B_j = \bigcup_{m \in M_j} C_m$
    for all $j \in J$.
    Let $M = \bigcup_{j \in J} M_j \subseteq \N$.
    For each $m \in M$, there exists $f(m) \in J$
    such that $m \in M_{f(m)}$.
    Then $\bigcup_{m \in M} B_{f(m)} = C_n$.
\end{proof}
\begin{remark}
    We don't actually need this.
\end{remark}

\begin{itemize}
    \item We use the same construction as in exercise 2 (a)
        on sheet 6.
        Let $A \subseteq X$ be analytic,
        i.e.~there exists a Polish space $Y$ and $f\colon Y \to X$ Borel
        with $f(Y) = X$.
        Then $f$ is still Borel with respect to the
        new topology, since Borel sets are preserved
        and by exercise 1 (b).
        % Let $(B_i)_{i < \omega}$ be a countable basis of $(X,\tau)$.
        % By a theorem from the lecture, there exists Polish
        % topologies $\cT_i$ such that $B_i$ is clopen wrt.~$\cT_i$
        % and $\cB(\cT_i) = \cB(\tau)$.
        % By a lemma from the lecture,
        % $\tau' \coloneqq \bigcup_i  \cT_i$
        % is Polish as well and $\cB(\tau') = \cB(\tau)$.
        % \todo{TODO: Basis}
    \item Suppose that there exist no disjoint clopen sets $U_0,U_1$,
        such that $W \cap U_0$ and $W \cap U_1$ are uncountable.

        Let $W_0 \coloneqq  W$
        Then there exist disjoint clopen sets $C_i^{(0)}$
        such that $W_0 \subseteq \bigcup_{i < \omega} C_i^{(0)}$
        and $\diam(C_i) < 1$,
        since $X$ is zero-dimensional.

        By assumption, exactly one of the $C_i^{(0)}$ has
        uncountable intersection with $W_0$.
        Let $i_0$  be such that  $W_0 \cap C_{i_0}^{(0)}$ is uncountable
        and set $W_1 \coloneqq W_0 \cap C_{i_0}^{(0)}$.
        Note that $W_0 \setminus W_1 = \bigcup_{i \neq i_0} C_i^{(0)}$ is countable.

        Let us recursively continue this construction:
        Suppose that $W_n$ uncountable has been chosen.
        Then choose $C_{i}^{(n)}$ clopen,
        disjoint with diameter $\le \frac{1}{n}$
        such that $W_n \subseteq \bigcup_{i} C_i^{(n)}$
        and let $i_n$ be the unique index
        such that $W_n \cap C_{i_n}^{(n)}$ is uncountable.

        Since $\diam(C_{i_n}^{(n)}) \xrightarrow{n \to \infty} 0$
        and the $C_{i_n}^{(n)}$ are closed,
        we get that $\bigcap_n C_{i_n}^{(n)}$
        contains exactly one point. Let that point be  $x$.

        However then
        \[
        W = \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right)
        \cup \bigcap_{n} (W \cap C_{i_n}^{(n)})
        = \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right) \cup \{x\}
        \]
        is countable as a countable union of countable sets $\lightning$.



        Other proof (without using the existence of a countable clopen
         basis):

        We can cover $X$ by countably many clopen sets of diameter
        $< \frac{1}{n}$:
        Cover $X$ with open balls of diameter $< \frac{1}{n}$.
        Write each open ball as a union of clopen sets.
        That gives us a cover by clopen sets of diameter $< \frac{1}{n}$.
        As $X$ is Lindelöf, there exists a countable subcover.
        Then continue as in the first proof.


    \item Note that this step does not help us to prove the statement.
        It was an error on the exercise sheet.

%         Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
%         as in the first part.
%         Clearly $f$ is also continuous with respect to the new topology,
%         so we may assume that $X$ is zero dimensional.
% 
%         Let $W \subseteq X$ be such that $f\defon{W}$ is injective
%         and $f(W) = f(X)$ (this exists by the axiom of choice).
%         Since $f(X)$ is uncountable, so is $W$.
%         By the second point, there exist disjoint clopen sets
%         $U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
%         are uncountable.
%         Inductively construct $U_s$ for $s \in 2^{<\omega}$
%         as follows:
%         Suppose that $U_{s}$ has already been chosen.
%         Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
%         be disjoint clopen such that $U_{s\concat 1} \cap W$
%         and $U_{s\concat  0} \cap W$ are uncountable.
%         Such sets exist, since $ U_s \cap W$ is uncountable
%         and $U_s$ is a zero dimensional space with the subspace topology.
%         And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
%         in $U_s$ iff it is clopen in $X$.

        Clearly this defines a Cantor scheme.
    \item Let $Y$ be a Polish space and $A \subseteq Y$ analytic
        and uncountable.
        Expand the topology on $Y$ so that $Y$ is zero dimensional
        and $A$ is still analytic.

        Then there exists a Polish space $X$
        and a continuous function $f\colon X \to Y$
        such that $f(X) = A$.

        $A$ is uncountable, so by (2)
        there exists non-empty disjoint clopen $V_0$, $V_1$
        such that $V_0 \cap A$ and $V_1 \cap A$
        are uncountable.

        Let $W_0 = f^{-1}(V_0 \cap A)$ and $W_1 = f^{-1}(V_1 \cap A)$.
        $ W_0$ and $W_1$ are clopen and disjoint.
        We can cover $W_0$ with countably many open sets
        of diameter $\le \frac{1}{n}$
        and similarly for $W_1$.
        Then pick open sets such that there image is uncountable.

        Repeating this construction we get a Cantor scheme on $X$.
        So we get $2^{\N} \overset{s}{\hookrightarrow} X$
        and by construction of the cantor scheme,
        we get that $f \circ s$ is injective and continuous.
\end{itemize}

\nr 4

Proof of Schröder-Bernstein:

Let $X_0 \coloneqq  X$, $Y_0 \coloneqq Y$
and define $X_{i+1} \coloneqq  g(Y_i)$, $Y_{i+1 } \coloneqq f(X_i)$.
We have $X_{i+1} \subseteq X_i$ and similarly for $Y_i$.
$f$ and $g$ are bijections between
$X_\omega \coloneqq \bigcap X_i$ and $Y_\omega \coloneqq \bigcap Y_i$.


% https://q.uiver.app/#q=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
\adjustbox{scale=0.7,center}{%
\begin{tikzcd}
	{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
	{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
	\arrow["f"'{pos=0.7}, from=1-2, to=2-4]
	\arrow["g"{pos=0.1}, from=2-2, to=1-4]
	\arrow["f"{pos=0.8}, from=1-6, to=2-8]
	\arrow["g"{pos=0.1}, from=2-6, to=1-8]
\end{tikzcd}
}

By \autoref{thm:lusinsouslin}
the injective image via a Borel set of a Borel set is Borel.

\autoref{thm:lusinsouslin} also gives that the inverse
of a bijective Borel map is Borel.
So we can just do the same proof and every set will be Borel.
