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inputs/lecture_15.tex Normal file
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@ -0,0 +1,3 @@
\lecture{15}{2023-12-05}{}
\todo{Lecture 15 is missing}

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inputs/tutorial_01.tex
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@ -1,6 +1,5 @@
\tutorial{01}{202-10-17}{} \tutorial{01}{2023-10-17}{}
% TODO MAIL
\begin{fact} \begin{fact}
A countable product of separable spaces $(X_n)_{n \in \N}$ is separable. A countable product of separable spaces $(X_n)_{n \in \N}$ is separable.
@ -66,5 +65,4 @@
In arbitrary topological spaces, In arbitrary topological spaces,
Lindelöf is the strongest of these notions. Lindelöf is the strongest of these notions.
\end{remark} \end{remark}

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inputs/tutorial_02.tex
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@ -1,8 +1,178 @@
\subsection{Sheet 1}
\tutorial{02}{2023-10-24}{} \tutorial{02}{2023-10-24}{}
% Points: 15 / 16 % Points: 15 / 16
\subsubsection{Exercise 4} \nr 1
\todo{handwritten solution}
\nr 2
\begin{enumerate}[(a)]
\item $d$ is an ultrametric:
Let $f,g,h \in X^{\N}$.
We need to show that $d(f,g) \le \max(d(f,h), d(g,h))$.
If $f = g$ this is trivial.
Otherwise let $n$ be minimal such that $f(n) \neq g(n)$.
Then $h(n) \neq f(n)$ or $ h(n) \neq g(n)$
must be the case.
W.l.o.g.~$h(n) \neq f(n)$.
Then $d(f,g) = \frac{1}{1+n} \le d(f,h)$.
\item $d$ induces the product topology on $X^{\N}$:
It suffices to show that the $\epsilon$-balls with respect to $d$
are exactly the basic open set of the product topology,
i.e.~the sets of the form
\[
\{x_1\} \times \ldots \times \{x_n\} \times X^{\N}
\]
for some $n \in \N$, $x_1,\ldots,x _n \in X$.
Let $\epsilon > 0$. Let $n$ be minimal such that $\frac{1}{1+n} \ge \epsilon$.
Then $B_{\epsilon}((x_i)_{i \in \N}) = \{x_1\} \times \{x_n\} \times X^{\N}$.
Since $\N \ni n \mapsto \frac{1}{1+n}$
is injective, every basic open set of the product topology
can be written in this way.
\item $d$ is complete:
Let $(f_n)_{n \in \N}$ be a Cauchy sequence with respect to $d$.
For $n \in \N$ take $N_n \in \N$
such that $d(f_i, f_j) < \frac{1}{1 + n}$.
Clearly $f_i(n) = f_j(n)$ for all $n > N_n$.
Define $f \in X^\N$ by $f(n) \coloneqq f_{N_n}(n)$.
Then $ (f_n)_{n \in \N}$
converges to $f$,
since for all $n > N_n$ $f_n$
\item If $X$ is countable, then $X^{\N}$ with the product topology
is a Polish space:
(We assume that $X$ is non-empty, as otherwise the claim is wrong)
We need to show that there exists a countable dense subset.
To this end, pick some $x_0 \in X$ and
consider the set $D \coloneqq \bigcup_{n\in \N} (X^n \times \{x_{0}\}^{\N})$.
Since $X$ is countable, so is $D$.
Take some $(a_n)_{n \in \N} \in X^{\N}$
and consider $B \coloneqq B_{\epsilon}((a_n)_{n \in \N})$.
Let $m$ be such that $\frac{1}{1+m} < \epsilon$.
Then $(b_{n})_{n \in \N} \in B \cap D$,
where $b_n \coloneqq a_n$ for $n \le m$ and $b_n \coloneqq x_0$
otherwise.
Hence $D$ is dense.
\end{enumerate}
\nr 3
\begin{enumerate}[(a)]
\item $S_{\infty}$ is a Polish space:
From (2) we know that $\N^{\N}$ is Polish.
Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
with respect to $\N^\N$.
Consider the sets $I \coloneqq \bigcap_{(i,j) \in \N^2, i < j} \{f \in \N^{\N} | f(i) \neq f(j)\}$
and $S \coloneqq \bigcap_{n \in \N} \{f \in \N^\N | n \in \im f\}$.
We have that $\{f \in \N^\N | f(i) \neq f(j)\} = \bigcup_{n \in \N} \N^{i-1} \times \{n\} \times \N^{i - j -1 } \times (\N \setminus \{n\} ) \times \N^\N$
is open.
Hence $I$ is $G_{\delta}$.
Furthermore $\{f \in \N^{\N} | n \in \im f\} = \bigcup_{k \in \N} \N^k \times \{n\} \times \N^\N$j
is open,
thus $S$ is $G_\delta$ as well.
In particular $S \cap G$ is $G_\delta$.
Since $I$ is the subset of injective functions
and $S$ is the subset of surjective functions,
we have that $S_{\infty} = I \cap S$.
\item $S_{\infty}$ is not locally compact:
Consider the point $x = (i)_{i \in \N} \in S_{\infty}$.
Let $x \in B$ be open. We need to show that there
is no closed compact set $C \supseteq B$
W.l.o.g.~let $B = (\{0\} \times \ldots \times \{n\} \times \N^\N) \cap S_\infty$
for some $n \in \N$.
Let $C \supseteq B$ be some closed set.
Consider the open covering
\[
\{S_{\infty} \setminus B\} \cup \{ B_j | j > n\}.
\]
where
\[
B_j \coloneqq (\{0\} \times \ldots \times \{n\} \times \{j\} \times \N^{\N}) \cap S_\infty.
\]
Clearly there cannot exist a finite subcover
as $B$ is the disjoint union of the $B_j$.
% TODO Think about this
\end{enumerate}
\nr 4
% (uniform metric)
%
% \begin{enumerate}[(a)]
% \item $d_u$ is a metric on $\cC(X,Y)$:
%
% It is clear that $d_u(f,f) = 0$.
%
% Let $f \neq g$. Then there exists $x \in X$ with
% $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$.
%
% Since $d$ is symmetric, so is $d_u$.
%
% Let $f,g,h \in \cC(X,Y)$.
% Take some $\epsilon > 0$
% choose $x_1, x_2 \in X$
% with $d_u(f,g) \le d(f(x_1), g(x_1)) + \epsilon$,
% $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$.
%
% Then for all $x \in X$
% \begin{IEEEeqnarray*}{rCl}
% d(f(x), h(x)) &\le &
% d(f(x), g(x)) + d(g(x), h(x))\\
% &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\
% &\le & d_u(f,g) + d_u(g,h) - 2\epsilon.
% \end{IEEEeqnarray*}
% Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$.
% Taking $\epsilon \to 0$ yields the triangle inequality.
%
% \item $\cC(X,Y)$ is a Polish space:
% \todo{handwritten solution}
%
% \begin{itemize}
% \item $d_u$ is a complete metric:
%
% Let $(f_n)_n$ be a Cauchy series with respect to $d_u$.
%
% Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect
% to $d$ for every $x$.
% Hence there exists a pointwise limit $f$ of the $f_n$.
% We need to show that $f$ is continuous.
%
% %\todo{something something uniform convergence theorem}
%
% \item $(\cC(X,Y), d_u)$ is separable:
%
% Since $Y$ is separable, there exists a countable
% dense subset $S \subseteq Y$.
%
% Consider $\cC(X,S) \subseteq \cC(X,Y)$.
% Take some $f \in \cC(X,Y)$.
% Since $X$ is compact,
%
%
% % TODO
%
% \end{itemize}
% \end{enumerate}
\begin{fact} \begin{fact}
Let $X $ be a compact Hausdorff space. Let $X $ be a compact Hausdorff space.
@ -32,7 +202,7 @@
Hausdorff spaces are normal Hausdorff spaces are normal
\end{proof} \end{proof}
Let $X$ be compact Polish (compact metrisable $\implies$ compact Polish) Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
and $Y $ Polish. and $Y $ Polish.
Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$. Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$. Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
@ -59,3 +229,4 @@ Clearly $d_u$ is a metric.
\begin{claim} \begin{claim}
There exists a countable dense subset. There exists a countable dense subset.
\end{claim} \end{claim}
\todo{handwritten solution}

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inputs/tutorial_03.tex
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@ -1,15 +1,61 @@
\ctutorial{03}{2023-10-31}{} \subsection{Sheet 2}
\tutorial{03}{2023-10-31}{}
% 15 / 16 % 15 / 16
\begin{remark} \begin{remark}
$F_\sigma$ $F_\sigma$ stands for \vocab{ferm\'e sum denumerable}.
stands for ferm\'e sum denumerable.
\end{remark} \end{remark}
\subsection{Exercise 2} \nr 1
Let $(U_i)_{i < \omega}$ be a countable base of $X$.
Define
\begin{IEEEeqnarray*}{rCl}
f\colon X &\longrightarrow & 2^{\omega} \\
x &\longmapsto & (x_i)_{i < \omega}
\end{IEEEeqnarray*}
where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
is open.
We have that $f$ is injective since $X$ is T1.
Let $f\colon X \hookrightarrow 2^\omega$ be such that
$f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\})$.
Let $V \subseteq 2^{\omega}$ be closed.
Then $2^{\omega} \setminus V$ is open, i.e.~has the form
$\bigcup_{i \in I} ((\prod_{j<n_j} X_{i,j}) \times 2^{\omega})$
for some $X_{i,j} \subseteq 2$.
As $2^{\omega}$ is second countable, we may assume $I$ to be countable.
Then $V = \bigcap_{i \in I} \left(2^{\omega} \setminus ((\prod_{i <n_j} X_{i,j}) \times 2^{\omega})\right)$.
Since $f$ is injective, we have $f^{-1}(\bigcap_{a \in A} a) = \bigcap_{a \in A} f^{-1}(a)$.
Thus it suffices to show that $f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
is $G_{\delta}$, as a countable intersection of $G_{\delta}$-sets
is $G_{\delta}$.
We have that $U_k \coloneqq f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 1\})$
is open. Since $f$ is injective
$f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 0\}) = X \setminus U_k$
is closed, in particular it is $G_\delta$.
Let $x = (x_1,\ldots, x_n) \in 2^{n} \setminus (\prod_{i < n} X_i)$.
Then $f^{-1}(\{x\} \times 2^\omega) = \bigcap_{i < n}\bigcap U'_n$
is $G_{\delta}$, were $U'_i = U_i$ if $x_k = i$
and $U'_i = X \setminus U_i$ otherwise.
Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$
is finite, we get that
$f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
is $G_\delta$ as a finite union of $G_{\delta}$ sets.
\nr 2
\todo{handwritten solution}
(b) (b)
Let $f(x^{(i)})$ be a sequence in $f(X)$. Let $f(x^{(i)})$ be a sequence in $f(X)$.
Suppose that $f(x^{(i)}) \to y$. Suppose that $f(x^{(i)}) \to y$.
@ -18,8 +64,7 @@ Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$. Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
\subsection{Exercise 3} \nr 3
\begin{example} \begin{example}
Consider Consider
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
@ -31,7 +76,6 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
and $\osc_f(x) = 0$ for $x \not\in \Q$. and $\osc_f(x) = 0$ for $x \not\in \Q$.
\end{example} \end{example}
\begin{definition} \begin{definition}
We say that $f\colon X \to Y$ is continuous We say that $f\colon X \to Y$ is continuous
at $a \in X$, at $a \in X$,
@ -40,13 +84,11 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
then $f^{-1}(N)$ is a neighbourhood of $a$. then $f^{-1}(N)$ is a neighbourhood of $a$.
\end{definition} \end{definition}
\begin{theorem}[Kuratowski] \begin{theorem}[Kuratowski]
Let $X$ be metrizable, $Y$ completely metrizable, Let $X$ be metrizable, $Y$ completely metrizable,
$f\colon S \to Y$ continuous. $S \subseteq X$ and $f\colon S \to Y$ continuous.
Then $f$ can be extended to a continuous fnuction $f_n$ Then $f$ can be extended to a continuous function $\tilde{f}$
on a $G_\delta$ set $G$ with $S \subseteq G \subseteq \overline{G}$. on a $G_\delta$ set $G$ with $S \subseteq G \subseteq \overline{S}$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
Let $G \coloneqq \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$. Let $G \coloneqq \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$.
@ -65,15 +107,12 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
is an intersection of open sets. is an intersection of open sets.
For $x \in G$, as $x \in \overline{S}$, For $x \in G$, as $x \in \overline{S}$,
there exists $(x_n)_{x_n < \omega}$, $x_n \in S_$ there exists $(x_n)_{x_n < \omega}$, $x_n \in S$
such that $x_n \to x$. such that $x_n \to x$.
We have that $(f(x_n))_n$ is Cauchy, We have that $(f(x_n))_n$ is Cauchy,
as $\osc_f(x) = 0$. as $\osc_f(x) = 0$.
% TODO
\todo{Something is missing here} \todo{Something is missing here}
\end{proof} \end{proof}
\begin{corollary} \begin{corollary}
@ -81,20 +120,74 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
Then $Y$ is $G_{\delta}$. Then $Y$ is $G_{\delta}$.
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Consider the identity $f\colon Y \hookrightarrow X$. \todo{TODO}
Then $f$ can be extended to a $G_{\delta}$ set. % Consider the identity $f\colon Y \hookrightarrow X$.
$f$ and $\id_G$ agree on $Y$. % Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
Hence $Y \subseteq G \subseteq \overline{Y}$. % with $Y \subseteq G \subseteq \overline{Y}$.
$Y$ is dense in $G$ and $\cod(f)$ is ltd.\todo{????} % $\tilde{f}$ and $\id_G$ agree on $Y$.
So $f = \id_G$, i.e.~$G = Y$. % $Y$ is dense in $G$ and the codomain of $f$ is ltd.\todo{????}
% So $f = \id_G$, i.e.~$G = Y$.
\end{proof} \end{proof}
\subsection{Exercise 4} \nr 4
\begin{enumerate}[(1)]
\item $f$ is a topological embedding:
Consider a basic open set
$B = \prod_{i < n} X_i \times \omega^{\omega}$
for some $X_i \subseteq \omega$.
Let $C$ be the subspace of $2^{\omega}$ consisting Then $f(B) = \left(\bigcup_{x \in \prod_{i<n} X_i} B_x \right) \cap f(\omega^{\omega})$
of sequences with finitely many $1$s. is open in $f(\omega^{\omega})$,
We want to show that $C \cong \Q$. where $B_x \coloneqq \{0^{x_0}10^{x_1}1\ldots 10^{x_n-1}1\} \times 2^{\omega}$.
On the other hand let $C = \{x_0x_1x_2x_3x_4 \ldots x_{n-1}\}\times 2^{\omega}$
be some basic open set of $2^{\omega}$.
W.l.o.g.~$x_0x_1\ldots x_{n-1}$
has the form $0^{a_0}10^{a_1}1\ldots 10^{a_k}x_{n-1}$.
If $x_{n-1} = 1$,
we get
\[
f^{-1}(C \cap f(\omega^{\omega})) = \{(a_0,a_1,\ldots,a_k)\} \times \omega^{\omega}.
\]
In the case of $x_{n-1} = 0$,
it is
\[
f^{-1}(C \cap f(\omega^\omega)) = \bigcup_{b > a_k} \{(a_0,a_1,\ldots,a_{k-1}, b)\} \times \omega^{\omega}.
\]
In both cases the preimage is open.
\item $C \coloneqq 2^{\omega} \setminus f(\omega^\omega)$
is countable and dense in $2^{\omega}$.
We have $C = \{x \in 2^{\omega} | x_i = 0 \text{ for all but finitely many $i$}\} = \bigcup_{i < \omega} (2^{i} \times 1^{\omega})$.
Clearly this is countable.
For denseness take some $x \in 2^\omega$.
Let $x^{(n)}$ be defined by $x^{(n)}_i = x_i$ for $i < n$
and $x^{(n)}_i = 0$ for $i \ge n$.
Then $x^{(n)} \in C$ for all $n$,
and $x^{(n)}$ converges to $x$.
\item $f(\omega^\omega)$ is $G_\delta$:
We have
\begin{IEEEeqnarray*}{rCl}
f(\omega^\omega) &=& 2^\omega \setminus \left(\bigcup_{i < \omega} (2^{i} \times 1^{\omega})\right)\\
&=& \bigcap_{i < \omega} \left(2^{\omega} \setminus(2^{i} \times 1^{\omega})\right).
\end{IEEEeqnarray*}
\item $C$ as in (2) is homeomorphic to $\Q$.
Go to the right in the even digits, go to the left for the odd digits, Go to the right in the even digits, go to the left for the odd digits,
i.e.~let $C = (1,-1,1,-1, \ldots)$ i.e.~let $C = (1,-1,1,-1, \ldots)$
and set $x < y$ iff $C \cdot x <_{\text{lex}} C \cdot y$. and set $x < y$ iff $C \cdot x <_{\text{lex}} C \cdot y$,
where $<_{\text{lex}}$ denotes the lexicographical ordering.
Note that the order topology of $<$ on $C$
agrees with the subspace topology from $2^\omega$.
By Cantor's theorem for countable, unbounded, dense linear
linear orders,
we get an order isomorphism $C \leftrightarrow \Q$.
This is also a homeomorphism, as the topologies
on $C$ and $\Q$ are the respective order topologies.
\end{enumerate}

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@ -1,70 +1,205 @@
\tutorial{04}{2023-11-14}{} \subsection{Sheet 3}
\subsection{Sheet 4} \tutorial{04}{}{}
% 14 / 20 \nr 1
\subsubsection{Exercise 1}
Let $A \neq \emptyset$ be discrete.
For $D \subseteq A^{\omega}$,
let
\[
T_D \coloneqq \{x\defon{n} \in A^{<\omega} | x \in D, n \in \N\}..
\]
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item $\langle X_\alpha : \alpha\rangle$ \item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:
is a descending chain of closed sets (transfinite induction).
Since $X$ is second countable, there cannot exist Clearly $T_D$ is a tree.
uncountable strictly decreasing chains of closed sets: Let $x \in T_D$.
Then there exists $d \in D$ such that $x = d\defon{n}$.
Hence $x \subseteq d\defon{n+1} \in T_D$.
Thus $x$ is not a leaf, i.e.~$T_D$ is pruned.
\item For any $T \subseteq A^{<\omega}$, $[T]$ is a closed subset
of $A^{\omega}$:
Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$ Let $a \in A^{\omega} \setminus [T]$.
was such a sequence, Then there exists some $n$ such that $a\defon{n} \not\in T$.
then $X \setminus X_{\alpha}$ is open for every $\alpha$, Hence $\{a_0\} \times \ldots \times \{a_{n+1}\} \times A^{\omega}$
Let $\{U_n : n < \omega\}$ be a countable basis. is an open neighbourhood of $a$ disjoint from $[T]$.
Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$,
is a strictly ascending chain in $\omega$.
\item We need to show that $X_{\alpha_0}$ is perfect and closed. \item $T \mapsto [T]$ is a bijection between the
It is closed since all $X_{\alpha}$ are, pruned trees on $A$ and the closed subsets of $A^{\omega}$.
and perfect, as a closed set $F$ is perfect
iff it coincides $F'$.
$X \setminus X_{\alpha_0}$ \begin{claim}
is countable: $[T_D] = D$ for any closed subset $D \subseteq A^{\omega}$.
$X_{\alpha} \setminus X_{\alpha + 1}$ is \end{claim}
countable as for every $x$ there exists a basic open set $U$, \begin{subproof}
such that $U \cap X_{\alpha} = \{x\}$, Clearly $D \subseteq [T_D]$.
and the space is second countable. Let $x \in [T_D]$.
Hence $X \setminus X_{\alpha_0}$ Then for every $n < \omega$,
is countable as a countable union of countable sets. there exists some $d_n \in D$ such that
$d_n\defon{n} = x\defon{n}$.
Clearly the $d_n$ converge to $x$.
Since $D$ is closed, we get $x \in D$.
\end{subproof}
This shows that $T \mapsto [T]$ is surjective.
Now let $T \neq T'$ be pruned trees.
Then there exists $x \in T \mathop{\triangle} T'$,
wlog.~$x \in T \setminus T'$.
Since $T$ is pruned
by applying the axiom of countable choice
we get an infinite branch $x' \in [T] \setminus [T']$.
Hence the map is injective.
\item Let $N_s \coloneqq \{x \in A^{\omega} | s \subseteq x\}$.
Show that every open $U \subseteq A^{\omega}$
can be written as $U = \bigcup_{s \in S} N_s$
for some set of pairwise incompatible $S \subseteq A^{<\omega}$.
Let $U$ be open.
Then $U$ has the form
\[
U = \bigcup_{i \in I} X_i \times A^{\omega}
\]
for some $X_i \subseteq A^{n_i}$, $n_i < \omega$.
Clearly
$U = \bigcup_{s \in S'} N_s$
for
$S' \coloneqq \bigcup_{i \in I} X_i$.
Define
\[
S \coloneqq \{s \in S' | \lnot\exists t \in S'.~t\subseteq s \land |t| < |s|\}.
\]
Then the elements of $S$ are pairwise incompatible
and $U = \bigcup_{s \in S} N_s$.
\item Let $T \subseteq A^{<\omega}$ be an infinite tree which is finitely
splitting.
Then $[T]$ is nonempty:
Let us recursively construct a sequence of compatible $s_n \in T$
with $|s_n| = n$
such that $\{s_n\} \times A^{<\omega} \cap T$ is infinite.
Let $s_0$ be the empty sequence;
by assumption $T$ is infinite.
Suppose that $s_n$ has been chosen.
Since $T$ is finitely splitting, there are only finitely
many $a \in A$ with $s_n\concat a \in T$.
Since $ \{s_n\} \times A^{<\omega} \cap T$
is infinite,
there must exist at least on $a \in A$
such that $\{s_n\concat a\} \times A^{<\omega} \cap T$
is infinite.
Define $s_{n+1} \coloneqq s_n \concat a$.
Then the union of the $s_n$ is an infinite branch of $T$,
i.e.~$[T]$ is nonempty.
\item Then $[T]$ is compact:
\todo{TODO}
% Let $\langle s_n, n <\omega \rangle$
% be a Cauchy sequence in $[T]$.
% Then for every $m < \omega$
% there exists an $N < \omega$ such that
% $s_n\defon{m} = s_{n'}\defon{m}$
% for all $n, n' > N$.
% Thus there exists a pointwise limit $s$ of the $s_n$.
% Since for all $m$ we have $s\defon{m} = s_n\defon{m} \in [T]$
% for $m$ large enough,
% we get $s \in [T]$.
% Hence $[T]$ is sequentially compact.
\end{enumerate} \end{enumerate}
\nr 2
\todo{handwritten}
\nr 3
\todo{handwritten}
\subsection{Exercise 3} \nr 4
\begin{itemize} \begin{notation}
\item Let $Y \subseteq \R$ be $G_\delta$ For $A \subseteq X$ let $A'$ denote the set of
such that $Y$ and $\R \setminus Y$ are dense in $\R$. accumulation points of $A$.
Then $Y \cong \cN$. \end{notation}
$Y$ is Polish, since it is $G_\delta$. \begin{theorem}
Let $X$ be a Polish space.
Then there exists a unique partition $X = P \sqcup U$
of $X$ into a perfect closed subset $P$ and a countable open subset $U$.
\end{theorem}
\begin{proof}
$Y$ is 0-dimensional, Let $P$ be the set of condensation points of $X$
since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$ and $U \coloneqq X \setminus P$.
form a clopen basis.
Each compact subset of $Y$ has empty interior: \begin{claim}
Let $K \subseteq Y$ be compact $U$ is open and countable.
and $U \subseteq K$ be open in $Y$. \end{claim}
Then we can find cover of $U$ that has no finite subcover $\lightning$. \begin{subproof}
Let $S$ be a countable dense subset.
For each $x \in U$,
there is an $\epsilon_x > 0$, $s_x \in S$ such that $x \in B_{\epsilon_x}(s_x)$
is at most countable.
Clearly $B_{\epsilon_x}(s_x) \subseteq U$,
as for every $y \in B_{\epsilon_x}(s_x)$,
$B_{\epsilon_x}(s_x)$ witnesses that $y \not\in P$.
Thus $U = \bigcup_{x \in U} B_{\epsilon_x}(s_x)$
is open.
Wlog.~$\epsilon_x \in \Q$ for all $x$.
Then the RHS is the union of at most
countably many countable sets, as $S \times \Q$
is countable.
\end{subproof}
\item Let $Y \subseteq \R$ be $G_\delta$ and dense \begin{claim}
such that $\R \setminus Y$ is dense as well. $P$ is perfect.
Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\} \subseteq \R^2$. \end{claim}
Then $Z$ is dense in $\R^2$ \begin{subproof}
and $\R^2 \setminus \Z$ is dense in $\R^2$. Let $x \in P'$ and $x \in U$ an open neighbourhood.
Then there exists $y \in P \cap U$.
In particular, $U$ is an open neighbourhood of $ y$,
hence $U$ is uncountable.
It follows that $x \in P$.
On the other hand let $x \in P$
and let $U$ be an open neighbourhood.
We need to show that $U \cap P \setminus \{x\}$
is not empty.
Suppose that for all $y \in U \cap P \setminus \{x\}$,
there is an open neighbourhood $U_y$
such that $U_y$ is at most countable.
Wlog.~$U_y = B_{\epsilon_y}(s_y)$ for some $s_y \in S$, $\epsilon_y > 0$,
where $S$ is again a countable dense subset.
Wlog.~$\epsilon_y \in \Q$.
But then
\[
U = \{x\} \cup \bigcup_{y \in U} B_{\epsilon_y}(s_y)
\]
is at most countable as a countable union of countable sets,
contradiction $x \in P$.
\end{subproof}
We have that for every $y \in Y$ \begin{claim}
$\partial B_y(0) \subseteq Z$. Let $P,U$ be defined as above
and let $P_2 \subseteq X$, $U_2 \subseteq X$
be such that $P_2$ is perfect and closed,
$U_2$ is countable and open
and $X = P_2 \sqcup U_2$.
Then $P_2 = P$ and $U_2 = U$.
\end{claim}
\todo{TODO}
\end{proof}
\begin{corollary}
Other example: Any Polish space is either countable or has cardinality equal to $\fc$.
Consider $\R^2 \setminus \Q^2$. \end{corollary}
\end{itemize} \begin{subproof}
\todo{TODO}
\end{subproof}

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\tutorial{05}{}{} \subsection{Sheet 4}
\tutorial{05}{2023-11-14}{}
% Sheet 5 - 18.5 / 20 % 14 / 20
\subsection{Exercise 1} \nr 1
Let $B \subseteq C$ be comeager.
Then $B = B_1 \cup B_2$,
where $B_1$ is dense $G_\delta$
and $B_2$ is meager.
\begin{fact} \begin{enumerate}[(a)]
$X$ is Baire iff every non-empty open set is non-meager. \item $\langle X_\alpha : \alpha\rangle$
is a descending chain of closed sets (transfinite induction).
In particular, let $X$ be Baire, Since $X$ is second countable, there cannot exist
then $U \overset{\text{open}}{\subseteq} X$ uncountable strictly decreasing chains of closed sets:
is Baire.
\end{fact}
\subsection{Exercise 4} Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$
was such a sequence,
then $X \setminus X_{\alpha}$ is open for every $\alpha$,
Let $\{U_n : n < \omega\}$ be a countable basis.
Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$,
is a strictly ascending chain in $\omega$.
\begin{enumerate}[(i)] \item We need to show that $X_{\alpha_0}$ is perfect and closed.
\item $|B| = \fc$, since $B$ contains a comeager It is closed since all $X_{\alpha}$ are,
$G_\delta$ set, $B'$: and perfect, as a closed set $F$ is perfect
$B'$ is Polish, iff it coincides $F'$.
hence $B' = P \cup C$
for $P$ perfect and $C$ countable, $X \setminus X_{\alpha_0}$
and $|P| \in \{\fc, 0\}$. is countable:
But $B'$ can't contain isolated point$. $X_{\alpha} \setminus X_{\alpha + 1}$ is
countable as for every $x$ there exists a basic open set $U$,
such that $U \cap X_{\alpha} = \{x\}$,
and the space is second countable.
Hence $X \setminus X_{\alpha_0}$
is countable as a countable union of countable sets.
\end{enumerate}
\nr 2
\todo{handwritten}
\nr 3
\begin{itemize}
\item Let $Y \subseteq \R$ be $G_\delta$
such that $Y$ and $\R \setminus Y$ are dense in $\R$.
Then $Y \cong \cN$.
$Y$ is Polish, since it is $G_\delta$.
$Y$ is 0-dimensional,
since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$
form a clopen basis.
Each compact subset of $Y$ has empty interior:
Let $K \subseteq Y$ be compact
and $U \subseteq K$ be open in $Y$.
Then we can find cover of $U$ that has no finite subcover $\lightning$.
\item Let $Y \subseteq \R$ be $G_\delta$ and dense
such that $\R \setminus Y$ is dense as well.
Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\} \subseteq \R^2$.
Then $Z$ is dense in $\R^2$
and $\R^2 \setminus \Z$ is dense in $\R^2$.
We have that for every $y \in Y$
$\partial B_y(0) \subseteq Z$.
Other example:
Consider $\R^2 \setminus \Q^2$.
\end{itemize}
\nr 4
\begin{enumerate}[(a)]
\item Let $d$ be a compatible, complete metric on $X$, wlog.~$d \le 1$.
Set $ U_{\emptyset} \coloneqq X$.
Suppose that $U_{s}$ has already been chosen.
Then $D_s \coloneqq X \setminus U_s$ is closed.
Hence $U_s^{(n)} \coloneqq \{x \in X | \dist(x,D_s) > \frac{1}{n}\}$
is open.
Let $m$ be such that $D_s^{(m)} \neq \emptyset$.
Clearly $\overline{U_s^{(n)}} \subseteq U_s$.
Let $(B_k)_{k < \omega}$ be a countable cover of $X$
consisting of balls of diameter $2^{-|s|-2}$.
Take some bijection $\phi\colon \omega \to \omega \times (\omega \setminus m)$
and set $U_{s\concat i} \coloneqq U_s^{(\pi_1\left( \phi(i) \right))} \cap B_{\pi_2(\phi(i))}$,
where there $\pi_i$ denote the projections
(if this is empty, set $U_{s \concat i} \coloneqq U_s^{\pi_1(\phi(j))} \cap B_{\pi_2(\phi(j))}$
for some arbitarily chosen $j < \omega$ such that it is not empty).
Then $\overline{U_{s \concat i}} \subseteq \overline{U_s^{(n)}} \subseteq U_s$,
\[
\bigcup_{i < \omega} U_{s \concat i} = \bigcup_{n < \omega} U_{s}^{\left( n \right) } = U_s
\]
and $\diam(U_{s \concat i}) \le \diam(B_{\pi_2(\phi(i))})$.
\item Let $s \in \omega^\omega$.
Then
\[
\bigcap_{n < \omega} \overline{U_{s\defon{n}}}
\]
contains exactly one point.
Let $f$ be the function that maps an $s \in \omega^{\omega}$
to the unique point in the intersection of the $\overline{U_{s\defon{n}}}$.
Let $x \in X$ be some point.
Then by induction we can construct a sequence $s \in \omega^{\omega}$
such that $x \in U_{s\defon{n}}$ for all $n$,
hence $x = f(s)$, i.e.~$f$ is surjective.
Let $B \overset{\text{open}}{\subseteq} X$.
Then $B = \bigcup_{i \in I} U_i$
for some $i \subseteq \omega^{<\omega}$,
as every basic open set can be recovered as a union of $U_i$
and $f^{-1}(B) = \bigcup_{i \in I} \left( \{i_0\} \times \ldots \{i_{|i|-1}\} \right) \times \omega^{\omega}$ is open,
hence $f$ is continuous.
On the other hand, consider an open ball $B \coloneqq \{\prod_{i < n} \{x_i\}\} \times \omega^{\omega} \subseteq \omega^{\omega}$.
Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open,
hence $f$ is open.
\end{enumerate} \end{enumerate}

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\tutorial{06}{2023-11-28}{} \subsection{Sheet 5}
% 5 / 20 \tutorial{06}{}{}
\subsection{Exercise 1} % Sheet 5 - 18.5 / 20
\begin{warning} \nr 1
Note that not every set has a density! \todo{handwritten}
\end{warning} Let $B \subseteq C$ be comeager.
Then $B = B_1 \cup B_2$,
\begin{enumerate}[(a)] where $B_1$ is dense $G_\delta$
\item Let $X = \bI^{\omega}$. and $B_2$ is meager.
Let $C_0 = \{(x_n) : x_n \to 0\}$.
Claim: $C_0 \in \Pi^0_3(X)$ (intersections of $F_\sigma$ sets).
We have
\[
x \in C_0 \iff \forall q \in \Q^+.~\exists N.~\forall n \ge N.~x_n \le q,
\]
i.e.
\[
C_0 = \bigcap_{q \in \Q^+}\bigcup_{N < \omega} \bigcap_{n > N} \{x_n : x_n \le q\}.
\]
Clearly this is a $\Pi^0_3$ set.
\item Let $Z \coloneqq \{f \in 2^{\omega} : f(\N) \text{ has density 0}\}$.
Claim: $Z \in \Pi^0_3(2^{\N})$.
It is
\[
Z = \bigcap_{q \in \Q^+} \bigcup_{N < \omega}
\bigcap_{n \ge N}\{f \in 2^{\omega} : \frac{\sum_{i < n} f(i)}{n} \le nq\}.
\]
Clearly this is a $\Pi^0_3$-set.
\end{enumerate}
\subsection{Exercise 2}
\begin{fact} \begin{fact}
Let $(X,\tau)$ be a Polish space and $X$ is Baire iff every non-empty open set is non-meager.
$A \in \cB(X)$.
Then there exists $\tau' \supseteq \tau$
with the same Borel sets as $\tau$
such that $A$ is clopen.
(Do it for $A$ closed, In particular, let $X$ be Baire,
then show that the sets which work then $U \overset{\text{open}}{\subseteq} X$
form a $\sigma$-algebra). is Baire.
\end{fact} \end{fact}
\nr 2
Let $(U_i)_{i < \omega}$ be a countable base of $Y$.
We want to find a $G_\delta$ set $A \subseteq X$
such that $f\defon{A}$ is continuous.
It suffices make sure that $f^{-1}\defon{A}(U_i)$ is open for all $i < \omega$.
Take some $i < \omega$.
Then $V_i \setminus M_i \subseteq f^{-1}(U_i) \subseteq V_i \cup M_i$,
where $V_i$ is open and $M_i$ is meager.
Let $M'_i \supseteq M_i$ be a meager $F_\sigma$-set.
Now let $A \coloneqq X \setminus \bigcup_{i <\omega} M_i'$.
We have that $A$ is a countable intersection of open dense sets,
hence it is dense and $G_\delta$.
For any $i < \omega$,
$V_i \cap A \subseteq f\defon{A}^{-1}(U_i) \subseteq (V_i \cup M_i) \cap A = V_i \cap A$,
so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
\nr 3
\todo{handwritten}
\nr 4
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item Let $(X, \tau)$ be Polish. \item $|B| = \fc$, since $B$ contains a comeager
We want to expand $\tau$ to a Polish topology $G_\delta$ set, $B'$:
$\tau_0$ maintaining the Borel sets, $B'$ is Polish,
such that $(X, \tau')$ is 0d. hence $B' = P \cup C$
for $P$ perfect and $C$ countable,
and $|P| \in \{\fc, 0\}$.
But $B'$ can't contain isolated point$.
\item To ensure that (a) holds, it suffices to chose
$a_i \not\in F_i$.
Since $|B| = \fc$ and $|\{a_i | j < i\}| = |i| < \fc$,
there exists some $x \in B \setminus \{\pi_1(a_j)| j <i\}$,
where $\pi$ denotes the projection.
Choose one such $x$.
We need to find $y \in \R$,
such that $(x,y) \not\in F_i$
and $\{a_j | j < i\} \cup \{(x,y)\}$
does not contain three collinear points.
Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$. Since $(F_i)_x$ is meager, we have that
Each $U_n$ is open, hence Borel, $|\{x\} \times \R \setminus F_i| = |\R \setminus (F_i)_x| = \fc$.
so by a theorem from the lecture$^{\text{tm}}$ Let $L \coloneqq \{y \in \R | \exists j < k < i. ~a_j, a_k,(x,y) \text{are collinear}\}$.
there exists a Polish topology $\tau_n$ Since every pair $a_j \neq a_k, j < k < i$,
such that $U_n$ is clopen, preserving Borel sets. adds at most one point to $L$,
we get $|L| \le |i|^2 < \fc$.
Hence $|\R \setminus (F_i)_x \setminus L| = \fc$.
In particular, the set is non empty, and we find $y$ as desired
and can set $a_i \coloneqq (x,y)$.
% We have not chosen too many points so far.
% So there are not too many lines,
% we can not choose a point from,
% but there are many points in $B$.
\item $A$ is by construction not a subset of any $F_\sigma$ meager set.
Hence it is not meager, since any meager set is contained
in an $F_\sigma$ meager set.
\item For every $x \in \R$ we have that $A_x$ contains at most
two points, hence it is meager.
In particular $\{x \in \R | A_x \text{ is meager}\} = \R$
is comeager.
However $A$ is not meager.
Hence $A$ can not be a set with the Baire property
by the theorem.
In particular, the assumption of the set having
the BP is necessary.
Hence we get $\tau_\infty$
such that all the $V_n$ are clopen in $\tau_\infty$.
Let $\tau^{1} \coloneqq \tau_\infty$.
Do this $\omega$-many times to get $\tau^{\omega}$.
$\tau^{\omega}$ has a base consisting
of finite intersections $A_1 \cap \ldots \cap A_n$,
where $A_i$ is a basis element we chose
to construct $\tau_i$,
hence clopen.
\item Let $(X, \tau_X), Y$ be Polish
and $f\colon X \to Y$ Borel.
Show $\exists \tau' \supseteq \tau$ maintaining the Borel structure
with $f$ continuous.
Let $(U_n)_n$ be a countable base of $Y$.
Clopenize all the preimages of the $(U_n)_n$.
\item Let $f\colon X \to Y$ be a Borel isomorphism.
Then there are finer topologies preserving the Borel
structure
such that $f\colon X' \to Y'$ is a homeomorphism.
Repeatedly apply (c).
Get $\tau_X^1$ to make $f$ continuous.
Then get $\tau_Y^1$ to make $f^{-1}$ continuous
(possibly violating continuity of $f$)
and so on.
Let $\tau_X^\omega \coloneqq \langle \tau_X^n \rangle$
and similarly for $\tau_Y^\omega$.
\end{enumerate}
\begin{idea}
If you do something and it didn't work,
try doing it again ($\omega$-many times).
\end{idea}
\subsection{Exercise 3}
\begin{enumerate}[(a)]
\item Show that if $\Gamma$ is self-dual (closed under complements)
and closed under continuous preimages,
then for any topological space $X$,
there does not exist an $X$-universal set for $\Gamma(X)$.
Suppose there is an $X$-universal set for $\Gamma(X)$,
i.e.~$U \subseteq X \times X$
such that $U \in \Gamma(X \times X) \land \{U_x : \in X\} = \Gamma(X)$.
Consider $X \xrightarrow[x\mapsto (x,x)]{d} X \times X$.
Let $V = U^c$.
Then $V \in \Gamma(X \times X)$ and $d^{-1}(V) \in \Gamma(X)$.
Then $d^{-1}(V) = U_x$ for some $x$.
But then $(x,x) \in U \iff x \in d^{-1}(V) \iff (x,x) \not\in U \lightning$.
\item Let $\xi$ be an ordinal
and let $X$ be a topological space.
Show that neither $\cB(X)$ nor $\Delta^0_\xi(X)$ can have $X$-universal
sets.
Clearly $\cB(X)$ is self-dual and closed under continuous preimages.
Clearly $\Delta^0_\xi(X)$ is self-dual
and closed under continuous preimages (by a trivial induction).
\end{enumerate} \end{enumerate}
\subsection{Exercise 4}
Recall:
\begin{fact}[Sheet 5, Exercise 1]
Let $\emptyset\neq X$ be a Baire space.
Then $\forall A \subseteq X$,
$A$ is either meager or locally comeager.
\end{fact}
\begin{theorem}[Kechris 16.1]
Let $X, Y$ be Polish.
Let
\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~
A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\]
Then $\cA$ contains all Borel sets.
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item Show for $V \in \cB(X), W \overset{\text{open}}{\subseteq} Y$
that $V \times W \in \cA$.
Clearly $V \times W$ is Borel
and $\{x \in X: W \cap U \text{ is not meager}\} \in \{\emptyset, V\}$.
\item Let $(A_n)_{n < \omega} \in \cA^{\omega}$.
Then $\bigcap_n A_n \in \cA$.
($(\bigcup_n A_n)_U = \bigcup_n (A_{n})_U$).
\item Let $A \in \cA$ and $B = A^c$.
Fix $\emptyset\neq U \subseteq Y$.
Then $\{x : A_x \text{is not meager in $U$}\}$ is Borel,
i.e.~$\{x : A_x^c \text{ is not meager in $U$}\}$ is Borel.
Since $A$ is Borel, $A_x$ is Borel as well.
Hence by the fact:
\begin{IEEEeqnarray*}{rCl}
&& \{x : A_x^c \text{ is not meager in $U$}\}\\
&=& \{x \colon A_x^c \text{ is locally comeager in $U$}\}\\
&=& \{x \colon \exists \emptyset\neq V \overset{\text{open}}{\subseteq} V.~ A_x \text{ is meager in $V$}\}\\
&=& \bigcup_{\emptyset \neq V \overset{\text{open}}{\subseteq} U} A_V^c
\end{IEEEeqnarray*}
(a countable union suffices, since we only need to check this for $V$ of the basis; if $A \subseteq V$ is nwd, then $A \cap U \subseteq U$ is nwd for all $U \overset{\text{open}}{\subseteq} V$).
\end{enumerate}
\end{proof}

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\tutorial{07}{2023-11-28}{}
% 5 / 20
\subsection{Exercise 1}
\begin{warning}
Note that not every set has a density!
\end{warning}
\begin{enumerate}[(a)]
\item Let $X = \bI^{\omega}$.
Let $C_0 = \{(x_n) : x_n \to 0\}$.
Claim: $C_0 \in \Pi^0_3(X)$ (intersections of $F_\sigma$ sets).
We have
\[
x \in C_0 \iff \forall q \in \Q^+.~\exists N.~\forall n \ge N.~x_n \le q,
\]
i.e.
\[
C_0 = \bigcap_{q \in \Q^+}\bigcup_{N < \omega} \bigcap_{n > N} \{x_n : x_n \le q\}.
\]
Clearly this is a $\Pi^0_3$ set.
\item Let $Z \coloneqq \{f \in 2^{\omega} : f(\N) \text{ has density 0}\}$.
Claim: $Z \in \Pi^0_3(2^{\N})$.
It is
\[
Z = \bigcap_{q \in \Q^+} \bigcup_{N < \omega}
\bigcap_{n \ge N}\{f \in 2^{\omega} : \frac{\sum_{i < n} f(i)}{n} \le nq\}.
\]
Clearly this is a $\Pi^0_3$-set.
\end{enumerate}
\subsection{Exercise 2}
\begin{fact}
Let $(X,\tau)$ be a Polish space and
$A \in \cB(X)$.
Then there exists $\tau' \supseteq \tau$
with the same Borel sets as $\tau$
such that $A$ is clopen.
(Do it for $A$ closed,
then show that the sets which work
form a $\sigma$-algebra).
\end{fact}
\begin{enumerate}[(a)]
\item Let $(X, \tau)$ be Polish.
We want to expand $\tau$ to a Polish topology
$\tau_0$ maintaining the Borel sets,
such that $(X, \tau')$ is 0d.
Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$.
Each $U_n$ is open, hence Borel,
so by a theorem from the lecture$^{\text{tm}}$
there exists a Polish topology $\tau_n$
such that $U_n$ is clopen, preserving Borel sets.
Hence we get $\tau_\infty$
such that all the $V_n$ are clopen in $\tau_\infty$.
Let $\tau^{1} \coloneqq \tau_\infty$.
Do this $\omega$-many times to get $\tau^{\omega}$.
$\tau^{\omega}$ has a base consisting
of finite intersections $A_1 \cap \ldots \cap A_n$,
where $A_i$ is a basis element we chose
to construct $\tau_i$,
hence clopen.
\item Let $(X, \tau_X), Y$ be Polish
and $f\colon X \to Y$ Borel.
Show $\exists \tau' \supseteq \tau$ maintaining the Borel structure
with $f$ continuous.
Let $(U_n)_n$ be a countable base of $Y$.
Clopenize all the preimages of the $(U_n)_n$.
\item Let $f\colon X \to Y$ be a Borel isomorphism.
Then there are finer topologies preserving the Borel
structure
such that $f\colon X' \to Y'$ is a homeomorphism.
Repeatedly apply (c).
Get $\tau_X^1$ to make $f$ continuous.
Then get $\tau_Y^1$ to make $f^{-1}$ continuous
(possibly violating continuity of $f$)
and so on.
Let $\tau_X^\omega \coloneqq \langle \tau_X^n \rangle$
and similarly for $\tau_Y^\omega$.
\end{enumerate}
\begin{idea}
If you do something and it didn't work,
try doing it again ($\omega$-many times).
\end{idea}
\subsection{Exercise 3}
\begin{enumerate}[(a)]
\item Show that if $\Gamma$ is self-dual (closed under complements)
and closed under continuous preimages,
then for any topological space $X$,
there does not exist an $X$-universal set for $\Gamma(X)$.
Suppose there is an $X$-universal set for $\Gamma(X)$,
i.e.~$U \subseteq X \times X$
such that $U \in \Gamma(X \times X) \land \{U_x : \in X\} = \Gamma(X)$.
Consider $X \xrightarrow[x\mapsto (x,x)]{d} X \times X$.
Let $V = U^c$.
Then $V \in \Gamma(X \times X)$ and $d^{-1}(V) \in \Gamma(X)$.
Then $d^{-1}(V) = U_x$ for some $x$.
But then $(x,x) \in U \iff x \in d^{-1}(V) \iff (x,x) \not\in U \lightning$.
\item Let $\xi$ be an ordinal
and let $X$ be a topological space.
Show that neither $\cB(X)$ nor $\Delta^0_\xi(X)$ can have $X$-universal
sets.
Clearly $\cB(X)$ is self-dual and closed under continuous preimages.
Clearly $\Delta^0_\xi(X)$ is self-dual
and closed under continuous preimages (by a trivial induction).
\end{enumerate}
\subsection{Exercise 4}
Recall:
\begin{fact}[Sheet 5, Exercise 1]
Let $\emptyset\neq X$ be a Baire space.
Then $\forall A \subseteq X$,
$A$ is either meager or locally comeager.
\end{fact}
\begin{theorem}[Kechris 16.1]
Let $X, Y$ be Polish.
Let
\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~
A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\]
Then $\cA$ contains all Borel sets.
\end{theorem}
\begin{proof}
\begin{enumerate}[(i)]
\item Show for $V \in \cB(X), W \overset{\text{open}}{\subseteq} Y$
that $V \times W \in \cA$.
Clearly $V \times W$ is Borel
and $\{x \in X: W \cap U \text{ is not meager}\} \in \{\emptyset, V\}$.
\item Let $(A_n)_{n < \omega} \in \cA^{\omega}$.
Then $\bigcap_n A_n \in \cA$.
($(\bigcup_n A_n)_U = \bigcup_n (A_{n})_U$).
\item Let $A \in \cA$ and $B = A^c$.
Fix $\emptyset\neq U \subseteq Y$.
Then $\{x : A_x \text{is not meager in $U$}\}$ is Borel,
i.e.~$\{x : A_x^c \text{ is not meager in $U$}\}$ is Borel.
Since $A$ is Borel, $A_x$ is Borel as well.
Hence by the fact:
\begin{IEEEeqnarray*}{rCl}
&& \{x : A_x^c \text{ is not meager in $U$}\}\\
&=& \{x \colon A_x^c \text{ is locally comeager in $U$}\}\\
&=& \{x \colon \exists \emptyset\neq V \overset{\text{open}}{\subseteq} V.~ A_x \text{ is meager in $V$}\}\\
&=& \bigcup_{\emptyset \neq V \overset{\text{open}}{\subseteq} U} A_V^c
\end{IEEEeqnarray*}
(a countable union suffices, since we only need to check this for $V$ of the basis; if $A \subseteq V$ is nwd, then $A \cap U \subseteq U$ is nwd for all $U \overset{\text{open}}{\subseteq} V$).
\end{enumerate}
\end{proof}