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\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 $\Sigma^1_1$:
% 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
\todo{Wait for mail}
\todo{Find a countable clopen base}
\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 $\lightning$.
\item 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 \todo{TODO}
\end{itemize}
\nr 4
Proof of Schröder-Bernstein:
Let $X_0 \coloneqq X$, $Y_0 \coloneqq Y$
and define $X_{i+1} \coloneqq g(Y_i)$, $Y_{i+1 } \coloneqq g(X_i)$.
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 $Y_\omega \coloneqq \bigcap Y_i$.
%\resizebox{\textwidth}{!}{%
% https://q.uiver.app/#q=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
\[\begin{tikzcd}
{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
\arrow["f"'{pos=0.7}, from=1-2, to=2-4]
\arrow["g"{pos=0.1}, from=2-2, to=1-4]
\arrow["f"{pos=0.8}, from=1-6, to=2-8]
\arrow["g"{pos=0.1}, from=2-6, to=1-8]
\end{tikzcd}\]
%}
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.

9
inputs/lecture_03.tex
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@ -150,9 +150,8 @@
\end{proof}
\begin{corollary}
\label{cor:perfectpolishcard}
Every nonempty perfect Polish
space $X$ has cardinality $\fc = 2^{\aleph_0}$
space $X$ has cardinality $C = 2^{\aleph_0}$
% TODO: eulerscript C ?
\end{corollary}
\begin{proof}
@ -164,11 +163,9 @@
\begin{theorem}
Any Polish space is countable
or it has cardinality $\fc$. % TODO C
or it has cardinality $C$. % TODO C
\end{theorem}
\begin{proof}
See \autoref{cor:polishcard}.
\end{proof}
\todo{Homework 3}
\begin{definition}

4
inputs/lecture_07.tex
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@ -1,5 +1,3 @@
\subsection{Sheet 6}
\lecture{07}{2023-11-07}{}
\begin{proposition}
@ -190,3 +188,5 @@
$\cB(\cT_\infty') = \cB(\cT)$
and $A $ is clopen in $\cT_{\infty}'$.
\end{refproof}

2
inputs/lecture_09.tex
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@ -119,7 +119,7 @@ We will see that not every analytic set is Borel.
Take $f^+\colon X \times Z \to Y \times Z, f^+ = f \times \id$.
Then
\[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
\item See \yaref{ex:7.2}.
\item \todo{Exercise}
\end{enumerate}
\end{proof}

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

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

177
inputs/tutorial_02.tex
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@ -1,178 +1,8 @@
\subsection{Sheet 1}
\tutorial{02}{2023-10-24}{}
% Points: 15 / 16
\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}
\subsubsection{Exercise 4}
\begin{fact}
Let $X $ be a compact Hausdorffspace.
@ -202,7 +32,7 @@
Hausdorff spaces are normal
\end{proof}
Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
Let $X$ be compact Polish (compact metrisable $\implies$ compact Polish)
and $Y $ Polish.
Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
@ -229,4 +59,3 @@ Clearly $d_u$ is a metric.
\begin{claim}
There exists a countable dense subset.
\end{claim}
\todo{handwritten solution}

147
inputs/tutorial_03.tex
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@ -1,61 +1,15 @@
\subsection{Sheet 2}
\tutorial{03}{2023-10-31}{}
\ctutorial{03}{2023-10-31}{}
% 15 / 16
\begin{remark}
$F_\sigma$ stands for \vocab{ferm\'e sum denumerable}.
$F_\sigma$
stands for ferm\'e sum denumerable.
\end{remark}
\nr 1
\subsection{Exercise 2}
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)
Let $f(x^{(i)})$ be a sequence in $f(X)$.
Suppose that $f(x^{(i)}) \to y$.
@ -64,7 +18,8 @@ Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
\nr 3
\subsection{Exercise 3}
\begin{example}
Consider
\begin{IEEEeqnarray*}{rCl}
@ -76,6 +31,7 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
and $\osc_f(x) = 0$ for $x \not\in \Q$.
\end{example}
\begin{definition}
We say that $f\colon X \to Y$ is continuous
at $a \in X$,
@ -84,11 +40,13 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
then $f^{-1}(N)$ is a neighbourhood of $a$.
\end{definition}
\begin{theorem}[Kuratowski]
Let $X$ be metrizable, $Y$ completely metrizable,
$S \subseteq X$ and $f\colon S \to Y$ continuous.
Then $f$ can be extended to a continuous function $\tilde{f}$
on a $G_\delta$ set $G$ with $S \subseteq G \subseteq \overline{S}$.
$f\colon S \to Y$ continuous.
Then $f$ can be extended to a continuous fnuction $f_n$
on a $G_\delta$ set $G$ with $S \subseteq G \subseteq \overline{G}$.
\end{theorem}
\begin{proof}
Let $G \coloneqq \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$.
@ -107,12 +65,15 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
is an intersection of open sets.
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 $x_n \to x$.
We have that $(f(x_n))_n$ is Cauchy,
as $\osc_f(x) = 0$.
% TODO
\todo{Something is missing here}
\end{proof}
\begin{corollary}
@ -120,74 +81,20 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
Then $Y$ is $G_{\delta}$.
\end{corollary}
\begin{proof}
\todo{TODO}
% Consider the identity $f\colon Y \hookrightarrow X$.
% Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
% with $Y \subseteq G \subseteq \overline{Y}$.
% $\tilde{f}$ and $\id_G$ agree on $Y$.
% $Y$ is dense in $G$ and the codomain of $f$ is ltd.\todo{????}
% So $f = \id_G$, i.e.~$G = Y$.
Consider the identity $f\colon Y \hookrightarrow X$.
Then $f$ can be extended to a $G_{\delta}$ set.
$f$ and $\id_G$ agree on $Y$.
Hence $Y \subseteq G \subseteq \overline{Y}$.
$Y$ is dense in $G$ and $\cod(f)$ is ltd.\todo{????}
So $f = \id_G$, i.e.~$G = Y$.
\end{proof}
\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 $X_i \subseteq \omega$.
\subsection{Exercise 4}
Then $f(B) = \left(\bigcup_{x \in \prod_{i<n} X_i} B_x \right) \cap f(\omega^{\omega})$
is open in $f(\omega^{\omega})$,
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 $\Q$.
Let $C$ be the subspace of $2^{\omega}$ consisting
of sequences with finitely many $1$s.
We want to show that $C \cong \Q$.
Go to the right in the even digits, go to the left for the odd digits,
i.e.~let $C = (1,-1,1,-1, \ldots)$
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}
and set $x < y$ iff $C \cdot x <_{\text{lex}} C \cdot y$.

240
inputs/tutorial_04.tex
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@ -1,208 +1,70 @@
\subsection{Sheet 3}
\tutorial{04}{2023-11-14}{}
\tutorial{04}{}{}
\subsection{Sheet 4}
\nr 1
% 14 / 20
\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)]
\item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:
\item $\langle X_\alpha : \alpha\rangle$
is a descending chain of closed sets (transfinite induction).
Clearly $T_D$ is a tree.
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 $A^{\omega}$:
Since $X$ is second countable, there cannot exist
uncountable strictly decreasing chains of closed sets:
Let $a \in A^{\omega} \setminus [T]$.
Then there exists some $n$ such that $a\defon{n} \not\in T$.
Hence $\{a_0\} \times \ldots \times \{a_{n+1}\} \times A^{\omega}$
is an open neighbourhood of $a$ disjoint from $[T]$.
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 $\omega$.
\item $T \mapsto [T]$ is a bijection between the
pruned trees on $A$ and the closed subsets of $A^{\omega}$.
\item We need to show that $X_{\alpha_0}$ is perfect and closed.
It is closed since all $X_{\alpha}$ are,
and perfect, as a closed set $F$ is perfect
iff it coincides $F'$.
\begin{claim}
$[T_D] = D$ for any closed subset $D \subseteq A^{\omega}$.
\end{claim}
\begin{subproof}
Clearly $D \subseteq [T_D]$.
Let $x \in [T_D]$.
Then for every $n < \omega$,
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.
$X \setminus X_{\alpha_0}$
is countable:
$X_{\alpha} \setminus X_{\alpha + 1}$ 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
\todo{handwritten}
\nr 4
\subsection{Exercise 3}
\begin{notation}
For $A \subseteq X$ let $A'$ denote the set of
accumulation points of $A$.
\end{notation}
\begin{itemize}
\item Let $Y \subseteq \R$ be $G_\delta$
such that $Y$ and $\R \setminus Y$ are dense in $\R$.
Then $Y \cong \cN$.
\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 Polish, since it is $G_\delta$.
Let $P$ be the set of condensation points of $X$
and $U \coloneqq X \setminus P$.
$Y$ is 0-dimensional,
since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$
form a clopen basis.
\begin{claim}
$U$ is open and countable.
\end{claim}
\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}
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 $\lightning$.
\begin{claim}
$P$ is perfect.
\end{claim}
\begin{subproof}
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 $x \in P$.
\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 $\R^2$.
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}
\begin{claim}
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}
We have that for every $y \in Y$
$\partial B_y(0) \subseteq Z$.
\begin{corollary}\label{cor:polishcard}
Any Polish space is either countable or has cardinality equal to $\fc$.
\end{corollary}
\begin{subproof}
Let $X = P \sqcup U$
where $P$ is perfect and $U$ is countable.
If $P \neq \emptyset$, we have $|P| = \fc$
by \yaref{cor:perfectpolishcard}.
\end{subproof}
Other example:
Consider $\R^2 \setminus \Q^2$.
\end{itemize}

138
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@ -1,123 +1,35 @@
\subsection{Sheet 4}
\tutorial{05}{2023-11-14}{}
\tutorial{05}{}{}
% 14 / 20
% Sheet 5 - 18.5 / 20
\nr 1
\subsection{Exercise 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{enumerate}[(a)]
\item $\langle X_\alpha : \alpha\rangle$
is a descending chain of closed sets (transfinite induction).
\begin{fact}
$X$ is Baire iff every non-empty open set is non-meager.
Since $X$ is second countable, there cannot exist
uncountable strictly decreasing chains of closed sets:
In particular, let $X$ be Baire,
then $U \overset{\text{open}}{\subseteq} X$
is Baire.
\end{fact}
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 $\omega$.
\subsection{Exercise 4}
\item We need to show that $X_{\alpha_0}$ is perfect and closed.
It is closed since all $X_{\alpha}$ are,
and perfect, as a closed set $F$ is perfect
iff it coincides $F'$.
$X \setminus X_{\alpha_0}$
is countable:
$X_{\alpha} \setminus X_{\alpha + 1}$ 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.
\begin{enumerate}[(i)]
\item $|B| = \fc$, since $B$ contains a comeager
$G_\delta$ set, $B'$:
$B'$ is Polish,
hence $B' = P \cup C$
for $P$ perfect and $C$ countable,
and $|P| \in \{\fc, 0\}$.
But $B'$ can't contain isolated point$.
\end{enumerate}

253
inputs/tutorial_06.tex
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@ -1,91 +1,180 @@
\subsection{Sheet 5}
\tutorial{06}{2023-11-28}{}
\tutorial{06}{}{}
% 5 / 20
% Sheet 5 - 18.5 / 20
\subsection{Exercise 1}
\nr 1
\todo{handwritten}
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{warning}
Note that not every set has a density!
\end{warning}
\begin{fact}
$X$ is Baire iff every non-empty open set is non-meager.
In particular, let $X$ be Baire,
then $U \overset{\text{open}}{\subseteq} X$
is Baire.
\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)]
\item $|B| = \fc$, since $B$ contains a comeager
$G_\delta$ set, $B'$:
$B'$ is Polish,
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.
\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).
Since $(F_i)_x$ is meager, we have that
$|\{x\} \times \R \setminus F_i| = |\R \setminus (F_i)_x| = \fc$.
Let $L \coloneqq \{y \in \R | \exists j < k < i. ~a_j, a_k,(x,y) \text{are collinear}\}$.
Since every pair $a_j \neq a_k, j < k < i$,
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.
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}

227
inputs/tutorial_07.tex
View file

@ -1,174 +1,93 @@
\subsection{Sheet 6}
\tutorial{07}{2023-11-28}{}
\tutorial{07}{2023-12-05}{}
% 17 / 20
% 5 / 20
\subsection{Exercise 2}
\nr 1
\begin{warning}
Note that not every set has a density!
\end{warning}
Recall \autoref{thm:analytic}.
\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).
Let $(A_i)_{i < \omega}$ be analytic subsets of a Polish space $X$.
We have
$\bigcap_i A_i$ is $\Sigma^1_1$:
% 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
\[
x \in C_0 \iff \forall q \in \Q^+.~\exists N.~\forall n \ge N.~x_n \le q,
D \coloneqq \{(y_n) : f_i(y_i) = f_j(y_j) ~ \forall i,j\}.
\]
i.e.
$D$ 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
\[
C_0 = \bigcap_{q \in \Q^+}\bigcup_{N < \omega} \bigcap_{n > N} \{x_n : x_n \le q\}.
C \coloneqq \{(x,(y^{(n)}) ) : \forall n.~(x, y^{(n)}) \in F_n\}.
\]
Clearly this is a $\Pi^0_3$ set.
$C$ is closed
and $\bigcap A_i = \proj_X(C)$.
\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 3}
\nr 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.
\begin{itemize}
\item Make $X$ zero dimensional preserving the Borel structure.
\item \todo{Find a countable clopen base}
\item
\end{itemize}
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}
\nr 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 $\Gamma(X)$.
\subsection{Exercise 4}
Suppose there is an $X$-universal set for $\Gamma(X)$,
i.e.~$U \subseteq X \times X$
such that $U \in \Gamma(X \times X) \land \{U_x : \in X\} = \Gamma(X)$.
Proof of Schröder-Bernstein:
Consider $X \xrightarrow[x\mapsto (x,x)]{d} X \times X$.
Let $X_0 \coloneqq X$, $Y_0 \coloneqq Y$
and define $X_{i+1} \coloneqq g(Y_i)$, $Y_{i+1 } \coloneqq g(X_i)$.
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 $Y_\omega \coloneqq \bigcap Y_i$.
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 $(x,x) \in U \iff x \in d^{-1}(V) \iff (x,x) \not\in U \lightning$.
% https://q.uiver.app/#q=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
\[\begin{tikzcd}
{X \setminus X_\omega} & {=} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
{Y\setminus Y_\omega} & {=} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
\arrow["f"'{pos=0.7}, from=1-3, to=2-5]
\arrow["g"{pos=0.1}, from=2-3, to=1-5]
\arrow["f"{pos=0.8}, from=1-7, to=2-9]
\arrow["g"{pos=0.1}, from=2-7, to=1-9]
\end{tikzcd}\]
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.
\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}
\nr 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}\footnote{See Kechris 16.1}
Let $X, Y$ be Polish.
For $\emptyset \neq U \overset{\text{open}}{\subseteq} Y$
let
\[A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\}.\]
Define
\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ A_U \text{ is Borel}\}.\]
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}

223
inputs/tutorial_08.tex
View file

@ -1,223 +0,0 @@
\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
\yalabel{Exercise}{}{ex:7.2}
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
\todo{Wait for mail}
\todo{Find a countable clopen base}
\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 $\lightning$.
\item 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 \todo{TODO}
\end{itemize}
\nr 4
Proof of Schröder-Bernstein:
Let $X_0 \coloneqq X$, $Y_0 \coloneqq Y$
and define $X_{i+1} \coloneqq g(Y_i)$, $Y_{i+1 } \coloneqq g(X_i)$.
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.

6
logic.sty
View file

@ -22,7 +22,6 @@
\usepackage{listings}
\usepackage{multirow}
\usepackage{float}
\usepackage{adjustbox}
\usepackage{quiver}
%\usepackage{algorithmicx}
@ -133,9 +132,6 @@
\DeclareSimpleMathOperator{LO} % linear orders
\DeclareSimpleMathOperator{WO} % well orderings
\DeclareSimpleMathOperator{osc} % oscillation
\newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}
%https://tex.stackexchange.com/questions/73437/how-do-i-typeset-the-concatenation-of-strings-properly
@ -147,5 +143,3 @@
\newcommand{\fc}{\mathfrak{c}}
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
\newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}
\newcommand\nr[1]{\subsubsection{Exercise #1}}

15
logic3.tex
View file

@ -15,7 +15,6 @@
\cleardoublepage
\setcounter{tocdepth}{2}
\tableofcontents
\cleardoublepage
@ -39,9 +38,6 @@
\input{inputs/lecture_12}
\input{inputs/lecture_13}
\input{inputs/lecture_14}
\input{inputs/lecture_15}
@ -49,17 +45,6 @@
\appendix
\section{Tutorial and Exercises}
\input{inputs/tutorial_01}
\input{inputs/tutorial_02}
\input{inputs/tutorial_03}
\input{inputs/tutorial_04}
\input{inputs/tutorial_05}
\input{inputs/tutorial_06}
\input{inputs/tutorial_07}
\input{inputs/tutorial_08}
\PrintVocabIndex