Josia Pietsch 3df55b6516
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\subsection{Sheet 7}
% 17 / 20
\nr 1
\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 $\zeta < \xi$.
\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)\}.
\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)\}.
\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 $f^{-1}(V) = \bigcup_n f^{-1}(V_n) \in \cB(X)$.
\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 $B_i \in \cB(Y_i)$.
\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,
$\coprod_{i < \omega} B_i \in \cB(Y)$
\[f(\coprod_{i < \omega} B_i) = \bigcup_i A_i.\]
\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 $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
\emph{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 $\bigcap A_i = \proj_X(C)$.
\nr 3
Let $X$ be a second-countable topological space.
Then every base of $X$ contains a countable subset which
is also a base of $X$.
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 $\bigcup_{m \in M} B_{f(m)} = C_n$.
We don't actually need this.
\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
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.
\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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
{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_1 \setminus X_2)} & \cup & {(X_2 \setminus X_3)} & \cdots & {} \\
{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_1 \setminus Y_2)} & \cup & {(Y_2 \setminus Y_3)} & \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]
By \autoref{thm:lusinsouslin}
the injective image via a Borel function 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.