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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=WzAsMTYsWzAsMCwiWCBcXHNldG1pbnVzIFhfXFxvbWVnYSA9Il0sWzAsMSwiWVxcc2V0bWludXMgWV9cXG9tZWdhID0iXSxbMSwwLCIoWF8wIFxcc2V0bWludXMgWF8xKSJdLFsxLDEsIihZXzAgXFxzZXRtaW51cyBZXzEpIl0sWzMsMSwiKFlfMCBcXHNldG1pbnVzIFlfMSkiXSxbNSwxLCIoWV8wIFxcc2V0bWludXMgWV8xKSJdLFszLDAsIihYXzAgXFxzZXRtaW51cyBYXzEpIl0sWzUsMCwiKFhfMCBcXHNldG1pbnVzIFhfMSkiXSxbNiwwLCJcXGNkb3RzIl0sWzYsMSwiXFxjZG90cyJdLFs0LDAsIlxcY3VwIl0sWzQsMSwiXFxjdXAiXSxbNywxXSxbNywwXSxbMiwwLCJcXGN1cCJdLFsyLDEsIlxcY3VwIl0sWzIsNCwiZiIsMix7ImxhYmVsX3Bvc2l0aW9uIjo3MH1dLFszLDYsImciLDAseyJsYWJlbF9wb3NpdGlvbiI6MTB9XSxbNywxMiwiZiIsMCx7ImxhYmVsX3Bvc2l0aW9uIjo4MH1dLFs1LDEzLCJnIiwwLHsibGFiZWxfcG9zaXRpb24iOjEwfV1d
\[\begin{tikzcd}
{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
\arrow["f"'{pos=0.7}, from=1-2, to=2-4]
\arrow["g"{pos=0.1}, from=2-2, to=1-4]
\arrow["f"{pos=0.8}, from=1-6, to=2-8]
\arrow["g"{pos=0.1}, from=2-6, to=1-8]
\end{tikzcd}\]
%}
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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@ -150,8 +150,9 @@
\end{proof}
\begin{corollary}
\label{cor:perfectpolishcard}
Every nonempty perfect Polish
space $X$ has cardinality $C = 2^{\aleph_0}$
space $X$ has cardinality $\fc = 2^{\aleph_0}$
% TODO: eulerscript C ?
\end{corollary}
\begin{proof}
@ -163,9 +164,11 @@
\begin{theorem}
Any Polish space is countable
or it has cardinality $C$. % TODO C
or it has cardinality $\fc$. % TODO C
\end{theorem}
\todo{Homework 3}
\begin{proof}
See \autoref{cor:polishcard}.
\end{proof}
\begin{definition}

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

3
inputs/lecture_09.tex
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@ -43,6 +43,7 @@ We will see that not every analytic set is Borel.
\end{remark}
\begin{theorem}
\label{thm:borel}
Let $X$ be Polish,
$\emptyset \neq A \subseteq X$.
Then the following are equivalent:
@ -118,7 +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 \todo{Exercise}
\item See \yaref{ex:7.2}.
\end{enumerate}
\end{proof}

2
inputs/lecture_12.tex
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@ -51,7 +51,7 @@
\begin{theorem}
\label{thm:lec12:1}
Suppose that $A \subseteq \cN$ is analytic.
Then there is $f\colon \cN \to \Tr$
Then there is $f\colon \cN \to \Tr$\todo{Borel?}
such that $x \in A \iff f(x)$ is ill-founded.
\end{theorem}
For the proof we need some prerequisites:

7
inputs/tutorial_04.tex
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@ -197,9 +197,12 @@ let
\todo{TODO}
\end{proof}
\begin{corollary}
\begin{corollary}\label{cor:polishcard}
Any Polish space is either countable or has cardinality equal to $\fc$.
\end{corollary}
\begin{subproof}
\todo{TODO}
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}

2
inputs/tutorial_06.tex
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@ -48,7 +48,7 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
hence $B' = P \cup C$
for $P$ perfect and $C$ countable,
and $|P| \in \{\fc, 0\}$.
But $B'$ can't contain isolated point$.
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 $|\{a_i | j < i\}| = |i| < \fc$,

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inputs/tutorial_07.tex
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@ -1,9 +1,9 @@
\subsection{Sheet 6}
\tutorial{07}{2023-11-28}{}
% 5 / 20
\subsection{Exercise 1}
\nr 1
\begin{warning}
Note that not every set has a density!
\end{warning}
@ -33,8 +33,7 @@
Clearly this is a $\Pi^0_3$-set.
\end{enumerate}
\subsection{Exercise 2}
\nr 2
\begin{fact}
Let $(X,\tau)$ be a Polish space and
$A \in \cB(X)$.
@ -96,8 +95,7 @@
try doing it again ($\omega$-many times).
\end{idea}
\subsection{Exercise 3}
\nr 3
\begin{enumerate}[(a)]
\item Show that if $\Gamma$ is self-dual (closed under complements)
and closed under continuous preimages,
@ -128,8 +126,7 @@
and closed under continuous preimages (by a trivial induction).
\end{enumerate}
\subsection{Exercise 4}
\nr 4
Recall:
\begin{fact}[Sheet 5, Exercise 1]
Let $\emptyset\neq X$ be a Baire space.
@ -137,12 +134,15 @@ Then $\forall A \subseteq X$,
$A$ is either meager or locally comeager.
\end{fact}
\begin{theorem}[Kechris 16.1]
\begin{theorem}\footnote{See Kechris 16.1}
Let $X, Y$ be Polish.
Let
\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~
A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\]
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}
@ -172,9 +172,3 @@ $A$ is either meager or locally comeager.
(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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@ -0,0 +1,223 @@
\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=WzAsMTYsWzAsMCwiWCBcXHNldG1pbnVzIFhfXFxvbWVnYSA9Il0sWzAsMSwiWVxcc2V0bWludXMgWV9cXG9tZWdhID0iXSxbMSwwLCIoWF8wIFxcc2V0bWludXMgWF8xKSJdLFsxLDEsIihZXzAgXFxzZXRtaW51cyBZXzEpIl0sWzMsMSwiKFlfMCBcXHNldG1pbnVzIFlfMSkiXSxbNSwxLCIoWV8wIFxcc2V0bWludXMgWV8xKSJdLFszLDAsIihYXzAgXFxzZXRtaW51cyBYXzEpIl0sWzUsMCwiKFhfMCBcXHNldG1pbnVzIFhfMSkiXSxbNiwwLCJcXGNkb3RzIl0sWzYsMSwiXFxjZG90cyJdLFs0LDAsIlxcY3VwIl0sWzQsMSwiXFxjdXAiXSxbNywxXSxbNywwXSxbMiwwLCJcXGN1cCJdLFsyLDEsIlxcY3VwIl0sWzIsNCwiZiIsMix7ImxhYmVsX3Bvc2l0aW9uIjo3MH1dLFszLDYsImciLDAseyJsYWJlbF9wb3NpdGlvbiI6MTB9XSxbNywxMiwiZiIsMCx7ImxhYmVsX3Bvc2l0aW9uIjo4MH1dLFs1LDEzLCJnIiwwLHsibGFiZWxfcG9zaXRpb24iOjEwfV1d
\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

@ -22,6 +22,7 @@
\usepackage{listings}
\usepackage{multirow}
\usepackage{float}
\usepackage{adjustbox}
\usepackage{quiver}
%\usepackage{algorithmicx}
@ -132,6 +133,9 @@
\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
@ -143,3 +147,5 @@
\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

@ -15,6 +15,7 @@
\cleardoublepage
\setcounter{tocdepth}{2}
\tableofcontents
\cleardoublepage
@ -38,6 +39,9 @@
\input{inputs/lecture_12}
\input{inputs/lecture_13}
\input{inputs/lecture_14}
\input{inputs/lecture_15}
@ -45,6 +49,17 @@
\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