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16
inputs/intro.tex
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@ -1,4 +1,6 @@
These are my notes on the lecture Probability Theory, \gist{%
These are my notes on the lecture
Logic 3: Abstract Topological Dynamics and Descriptive Set Theory
taught by \textsc{Aleksandra Kwiatkowska} taught by \textsc{Aleksandra Kwiatkowska}
in the summer term 2023 at the University Münster. in the summer term 2023 at the University Münster.
@ -18,3 +20,15 @@ I could not attend!
This notes follow the way the material was presented in the lecture rather This notes follow the way the material was presented in the lecture rather
closely. Additions (e.g.~from exercise sheets) closely. Additions (e.g.~from exercise sheets)
and slight modifications have been marked with $\dagger$. and slight modifications have been marked with $\dagger$.
}{
This document aims to give a very brief summary of
my \href{https://josia-notes.users.abstractnonsen.se/w23-logic-3/logic3.pdf}{notes on the course Logic 3}.
I try to omit most technical details and only summarize the most important
ideas.
Note that this is currently work in progress.
Currently the differences to the original document are only
minor (this is still mostly a technical test),
but this document will get shorter as I work through
and summarize it.
}

29
inputs/lecture_01.tex
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@ -58,7 +58,7 @@ However the converse of this does not hold.
Take $x_0 \in X$ and consider the topology given by Take $x_0 \in X$ and consider the topology given by
\[ \[
\tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}. \tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}.
\] \]
Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable. Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
\end{example} \end{example}
\begin{example}[Sorgenfrey line] \begin{example}[Sorgenfrey line]
@ -76,7 +76,7 @@ However the converse of this does not hold.
\end{itemize} \end{itemize}
\end{fact} \end{fact}
\begin{fact} \begin{fact}
Compact\footnote{It is not clear whether compact means compact and Hausdorff in this lecture.} Hausdorff spaces are \vocab{normal} (T4) Compact Hausdorff spaces are \vocab{normal} (T4)
i.e.~two disjoint closed subsets can be separated i.e.~two disjoint closed subsets can be separated
by open sets. by open sets.
\end{fact} \end{fact}
@ -114,7 +114,7 @@ However the converse of this does not hold.
\end{absolutelynopagebreak} \end{absolutelynopagebreak}
\subsection{Some facts about polish spaces} \subsection{Some facts about polish spaces}
\gist{%
\begin{fact} \begin{fact}
Let $(X, \tau)$ be a topological space. Let $(X, \tau)$ be a topological space.
Let $d$ be a metric on $X$. Let $d$ be a metric on $X$.
@ -130,9 +130,10 @@ To show that $\tau_d = \tau_{d'}$
for two metrics $d, d'$, for two metrics $d, d'$,
suffices to show that open balls in one metric are unions of open balls in the other. suffices to show that open balls in one metric are unions of open balls in the other.
\end{fact} \end{fact}
}{}
\begin{notation} \begin{notation}
We sometimes denote $\min(a,b)$ by $a \wedge b$. We sometimes\footnote{only in this subsection?} denote $\min(a,b)$ by $a \wedge b$.
\end{notation} \end{notation}
\begin{proposition} \begin{proposition}
@ -142,18 +143,20 @@ suffices to show that open balls in one metric are unions of open balls in the o
Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$. Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
\end{proposition} \end{proposition}
\gist{%
\begin{proof} \begin{proof}
To check the triangle inequality: To check the triangle inequality:
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right) \wedge 1\\ d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right) \wedge 1\\
&\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right). &\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right).
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$ For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$
and for $\epsilon > 1$, $B'_\epsilon(x) = X$. and for $\epsilon > 1$, $B'_\epsilon(x) = X$.
Since $d$ is complete, we have that $d'$ is complete. Since $d$ is complete, we have that $d'$ is complete.
\end{proof} \end{proof}
}{}
\begin{proposition} \begin{proposition}
Let $A$ be a Polish space. Let $A$ be a Polish space.
Then $A^{\omega}$ Polish. Then $A^{\omega}$ Polish.
@ -164,11 +167,11 @@ suffices to show that open balls in one metric are unions of open balls in the o
(Consider the basic open sets of the product topology). (Consider the basic open sets of the product topology).
Let $d \le 1$ be a complete metric on $A$. Let $d \le 1$ be a complete metric on $A$.
Define $D$ on $A^\omega$ by Define $D$ on $A^\omega$ by
\[ \[
D\left( (x_n), (y_n) \right) \coloneqq D\left( (x_n), (y_n) \right) \coloneqq
\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n). \sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
\] \]
Clearly $D \le 1$. Clearly $D \le 1$.
It is also clear, that $D$ is a metric. It is also clear, that $D$ is a metric.
@ -194,7 +197,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}. Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
Then $X$ topologically embeds into the Then $X$ topologically embeds into the
\vocab{Hilbert cube}, \vocab{Hilbert cube},
i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$ i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$
such that $f: X \to f(X)$ is a homeomorphism. such that $f: X \to f(X)$ is a homeomorphism.
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
@ -252,15 +255,15 @@ suffices to show that open balls in one metric are unions of open balls in the o
\begin{proposition} \begin{proposition}
Closed subspaces of Polish spaces are Polish. Closed subspaces of Polish spaces are Polish.
\end{proposition} \end{proposition}
\gist{}{ \gist{%
\begin{proof} \begin{proof}
Let $X$ be Polish and $V \subseteq X$ closed. Let $X$ be Polish and $V \subseteq X$ closed.
Let $d$ be a complete metric on $X$. Let $d$ be a complete metric on $X$.
Then $d\defon{V}$ is complete. Then $d\defon{V}$ is complete.
Subspaces of second countable spaces Subspaces of second countable spaces
are second countable. are second countable.
\end{proof} \end{proof}%
} }{}
\begin{definition} \begin{definition}
Let $X$ be a topological space. Let $X$ be a topological space.

117
inputs/lecture_02.tex
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@ -72,41 +72,51 @@
\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\] \[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
metric is complete. metric is complete.
$f_U$ is an embedding of $U$ into $X \times \R$\gist{: $f_U$ is an embedding of $U$ into $X \times \R$%
\begin{itemize} \gist{:
\item It is injective because of the first coordinate. \begin{itemize}
\item It is continuous since $d(x, U^c)$ is continuous \item It is injective because of the first coordinate.
and only takes strictly positive values. % TODO \item It is continuous since $d(x, U^c)$ is continuous
\item The inverse is continuous because projections and only takes strictly positive values. % TODO
are continuous. \item The inverse is continuous because projections
\end{itemize} are continuous.
}{.} \end{itemize}
}{.}
So we have shown that $U$ and \gist{%
the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$ So we have shown that $U$ and
are homeomorphic. the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
The graph is closed \gist{in $U \times \R$, are homeomorphic.
because $\tilde{f_U}$ is continuous. The graph is closed \gist{in $U \times \R$,
It is closed}{} in $X \times \R$ \gist{because because $\tilde{f_U}$ is continuous.
$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}. It is closed}{} in $X \times \R$ \gist{because
\todo{Make this precise} $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
\todo{Make this precise}
Therefore we identified $U$ with a closed subspace of
the Polish space $(X \times \R, d_1)$.
}{%
So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$
and the RHS is a close subspace of the Polish space
$(X \times \R, d_1)$.
}
Therefore we identified $U$ with a closed subspace of
the Polish space $(X \times \R, d_1)$.
\end{refproof} \end{refproof}
Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$. Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
Take Consider
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\ f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
x &\longmapsto & x &\longmapsto &
\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right) \left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
As for an open $U$, $f_Y$ is an embedding. \gist{
Since $X \times \R^{\N}$ As for an open $U$, $f_Y$ is an embedding.
is completely metrizable, Since $X \times \R^{\N}$
so is the closed set $f_Y(Y) \subseteq X \times \R^\N$. is completely metrizable,
so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
}{}
\begin{claim} \begin{claim}
\label{psubspacegdelta:c2} \label{psubspacegdelta:c2}
@ -123,36 +133,35 @@
\item $\diam_d(U) \le \frac{1}{n}$, \item $\diam_d(U) \le \frac{1}{n}$,
\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$. \item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
\end{enumerate} \end{enumerate}
\gist{ \gist{%
We want to show that $Y = \bigcap_{n \in \N} V_n$. We want to show that $Y = \bigcap_{n \in \N} V_n$.
For $x \in Y$, $n \in \N$ we have $x \in V_n$, For $x \in Y$, $n \in \N$ we have $x \in V_n$,
as we can choose two neighbourhoods as we can choose two neighbourhoods
$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$, $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
such that $\diam_{d_Y}(U) < \frac{1}{n}$ such that $\diam_{d_Y}(U) < \frac{1}{n}$
and $U_2 \cap Y = U_1$. and $U_2 \cap Y = U_1$.
Additionally choose $x \in U_3$ open in $X$ Additionally choose $x \in U_3$ open in $X$
with $\diam_{d}(U_3) < \frac{1}{n}$. with $\diam_{d}(U_3) < \frac{1}{n}$.
Then consider $U_2 \cap U_3 \subseteq V_n$. Then consider $U_2 \cap U_3 \subseteq V_n$.
Hence $Y \subseteq \bigcap_{n \in \N} V_n$. Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
Now let $x \in \bigcap_{n \in \N} V_n$. Now let $x \in \bigcap_{n \in \N} V_n$.
For each $n$ pick $x \in U_n \subseteq X$ open For each $n$ pick $x \in U_n \subseteq X$ open
satisfying (i), (ii), (iii). satisfying (i), (ii), (iii).
From (i) and (ii) it follows that $x \in \overline{Y}$, From (i) and (ii) it follows that $x \in \overline{Y}$,
since we can consider a sequence of points $y_n \in U_n \cap Y$ since we can consider a sequence of points $y_n \in U_n \cap Y$
and get $y_n \xrightarrow{d} x$. and get $y_n \xrightarrow{d} x$.
For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$ For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
is an open set containing $x$, is an open set containing $x$,
hence $U_n' \cap Y \neq \emptyset$. hence $U_n' \cap Y \neq \emptyset$.
Thus we may assume that the $U_i$ form a decreasing sequence. Thus we may assume that the $U_i$ form a decreasing sequence.
We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$. We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$, If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$ since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$. and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
The sequence $y_n$ converges to the unique point in The sequence $y_n$ converges to the unique point in
$\bigcap_{n} \overline{U_n \cap Y}$. $\bigcap_{n} \overline{U_n \cap Y}$.
Since the topologies agree, this point is $x$. Since the topologies agree, this point is $x$.
}{Then $Y = \bigcap_n U_n$.} }{Then $Y = \bigcap_n U_n$.}
\end{refproof} \end{refproof}
\end{refproof} \end{refproof}

86
inputs/lecture_03.tex
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@ -2,7 +2,7 @@
\subsection{Trees} \subsection{Trees}
\gist{%
\begin{notation} \begin{notation}
Let $A \neq \emptyset$, $n \in \N$. Let $A \neq \emptyset$, $n \in \N$.
Then Then
@ -34,7 +34,7 @@
$s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$. $s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$.
Let $s,t \in A^{<\N}$. Let $s,t \in A^{<\N}$.
We say that $s$ is an \vocab{initial segment} We say that $s$ is an \vocab{initial segment}
of $t$ (or $t$ is an \vocab{extension} of $s$) of $t$ (or $t$ is an \vocab{extension} of $s$)
if there exists an $n$ such that $s = t\defon{|s|}$. if there exists an $n$ such that $s = t\defon{|s|}$.
We write this as $s \subseteq t$. We write this as $s \subseteq t$.
@ -44,8 +44,8 @@
Otherwise the are \vocab{incompatible}, Otherwise the are \vocab{incompatible},
we denote that as $s \perp t$. we denote that as $s \perp t$.
The \vocab{concatenation} The \vocab{concatenation}
of $s = (s_0,\ldots, s_{n-1})$ of $s = (s_0,\ldots, s_{n-1})$
and $t = (t_0,\ldots, t_{m-1})$ and $t = (t_0,\ldots, t_{m-1})$
is the sequence is the sequence
$s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$ $s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$
@ -59,10 +59,11 @@
define extension, initial segments define extension, initial segments
and concatenation of a finite sequence with an infinite one. and concatenation of a finite sequence with an infinite one.
\end{notation} \end{notation}
}{}
\begin{definition} \begin{definition}
A \vocab{tree} A \vocab{tree}
on a set $A$ is a subset $T \subseteq A^{<\N}$ on a set $A$ is a subset $T \subseteq A^{<\N}$
closed under initial segments, closed under initial segments,
i.e.~if $t \in T, s \subseteq t \implies s \in T$. i.e.~if $t \in T, s \subseteq t \implies s \in T$.
Elements of trees are called \vocab{nodes}. Elements of trees are called \vocab{nodes}.
@ -70,27 +71,27 @@
A \vocab{leave} is an element of $T$, A \vocab{leave} is an element of $T$,
that has no extension in $t$. that has no extension in $t$.
An \vocab{infinite branch} of a tree $T$ An \vocab{infinite branch} of a tree $T$
is $x \in A^{\N}$ is $x \in A^{\N}$
such that $\forall n.~x\defon{n} \in T$. such that $\forall n.~x\defon{n} \in T$.
The \vocab{body} of $T$ is the set of all The \vocab{body} of $T$ is the set of all
infinite branches: infinite branches:
\[ \[
[T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}. [T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.
\] \]
We say that $T$ is \vocab{pruned}, We say that $T$ is \vocab{pruned},
iff iff
\[ \[
\forall t\in T.\exists s \supsetneq t.~s \in T. \forall t\in T.\exists s \supsetneq t.~s \in T.
\] \]
\end{definition} \end{definition}
\begin{definition} \begin{definition}
A \vocab{Cantor scheme} A \vocab{Cantor scheme}
on a set $X$ is a family on a set $X$ is a family
$(A_s)_{s \in 2^{< \N}}$ $(A_s)_{s \in 2^{< \N}}$
of subsets of $X$ such that of subsets of $X$ such that
\begin{enumerate}[i)] \begin{enumerate}[i)]
\item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$. \item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
@ -99,7 +100,7 @@
\end{definition} \end{definition}
\begin{definition} \begin{definition}
A topological space A topological space
is \vocab{perfect} is \vocab{perfect}
if it has no isolated points, if it has no isolated points,
i.e.~for any $U \neq \emptyset$ open, i.e.~for any $U \neq \emptyset$ open,
@ -108,15 +109,15 @@
\begin{theorem} \begin{theorem}
\label{thm:cantortopolish} \label{thm:cantortopolish}
Let $X \neq \emptyset$ Let $X \neq \emptyset$
be a perfect Polish space. be a perfect Polish space.
Then there is an embedding Then there is an embedding
of the Cantor space $2^{\N}$ of the Cantor space $2^{\N}$
into $X$. into $X$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
We will define a Cantor scheme We will define a Cantor scheme
$(U_s)_{s \in 2^{<\N}}$ $(U_s)_{s \in 2^{<\N}}$
such that $\forall s \in 2^{< \N}$. such that $\forall s \in 2^{< \N}$.
\begin{enumerate}[(i)] \begin{enumerate}[(i)]
\item $U_s \neq \emptyset$ and open, \item $U_s \neq \emptyset$ and open,
@ -127,39 +128,46 @@
We define $U_s$ inductively on the length of $s$. We define $U_s$ inductively on the length of $s$.
For $U_{\emptyset}$ take any non-empty open set \gist{%
with small enough diameter. For $U_{\emptyset}$ take any non-empty open set
with small enough diameter.
Given $U_s$, pick $x \neq y \in U_s$ Given $U_s$, pick $x \neq y \in U_s$
and let $U_{s \concat 0} \ni x$, and let $U_{s \concat 0} \ni x$,
$U_{s \concat 1} \ni y$ $U_{s \concat 1} \ni y$
be disjoint, open, be disjoint, open,
of diameter $\le \frac{1}{2^{|s| +1}}$ of diameter $\le \frac{1}{2^{|s| +1}}$
and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$. and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
}{}
\gist{%
Let $x \in 2^{\N}$. Let $x \in 2^{\N}$.
Then let $f(x)$ be the unique point in $X$ Then let $f(x)$ be the unique point in $X$
such that such that
\[ \[
\{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}. \{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}.
\] \]
(This is nonempty as $X$ is a completely metrizable space.) (This is nonempty as $X$ is a completely metrizable space.)
It is clear that $f$ is injective and continuous. It is clear that $f$ is injective and continuous.
% TODO: more details % TODO: more details
$2^{\N}$ is compact, hence $f^{-1}$ is also continuous. $2^{\N}$ is compact, hence $f^{-1}$ is also continuous.
}{Consider $f\colon 2^{\N} \hookrightarrow X, x \mapsto y$, where $\{y\} = \bigcap_n U_{x\defon n}$.
By compactness of $2^{\N}$, we get that $f^{-1}$ is continuous.}
\end{proof} \end{proof}
\begin{corollary} \begin{corollary}
\label{cor:perfectpolishcard} \label{cor:perfectpolishcard}
Every nonempty perfect Polish Every nonempty perfect Polish
space $X$ has cardinality $\fc = 2^{\aleph_0}$ space $X$ has cardinality $\fc = 2^{\aleph_0}$
% TODO: eulerscript C ?
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Since the cantor space embeds into $X$, \gist{%
we get the lower bound. Since the cantor space embeds into $X$,
Since $X$ is second countable and Hausdorff, we get the lower bound.
we get the upper bound. Since $X$ is second countable and Hausdorff,
we get the upper bound.%
}{Lower bound: $2^{\N} \hookrightarrow X$,
upper bound: \nth{2} countable and Hausdorff.}
\end{proof} \end{proof}
\begin{theorem} \begin{theorem}
@ -173,13 +181,13 @@
\begin{definition} \begin{definition}
A \vocab{Lusin scheme} on a set $X$ A \vocab{Lusin scheme} on a set $X$
is a family $(A_s)_{s \in \N^{<\N}}$ is a family $(A_s)_{s \in \N^{<\N}}$
of subsets of $X $ of subsets of $X $
such that such that
\begin{enumerate}[(i)] \begin{enumerate}[(i)]
\item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$ \item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$
for all $j \neq i \in \N$, $s \in \N^{<\N}$. for all $j \neq i \in \N$, $s \in \N^{<\N}$.
\item $A_{s \concat i} \subseteq A_s$ \item $A_{s \concat i} \subseteq A_s$
for all $i \in \N, s \in \N^{<\N}$. for all $i \in \N, s \in \N^{<\N}$.
\end{enumerate} \end{enumerate}
\end{definition} \end{definition}
@ -191,11 +199,11 @@
\[ \[
D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN
% TODO correct N for the Baire space? % TODO correct N for the Baire space?
\] \]
and a continuous bijection from and a continuous bijection from
$D$ onto $X$ (the inverse does not need to be continuous). $D$ onto $X$ (the inverse does not need to be continuous).
Moreover there is a continuous surjection $g: \cN \to X$ Moreover there is a continuous surjection $g: \cN \to X$
extending $f$. extending $f$.
\end{theorem} \end{theorem}
\begin{definition} \begin{definition}
@ -203,12 +211,14 @@
countable union of closed sets, countable union of closed sets,
i.e.~the complement of a $G_\delta$ set. i.e.~the complement of a $G_\delta$ set.
\end{definition} \end{definition}
\gist{%
\begin{observe} \begin{observe}
\begin{itemize} \begin{itemize}
\item Any open set is $F {\sigma}$. \item Any open set is $F {\sigma}$.
\item In metric spaces the intersection of an open and closed set is $F_\sigma$. \item In metric spaces the intersection of an open and closed set is $F_\sigma$.
\end{itemize} \end{itemize}
\end{observe} \end{observe}
}{}
\begin{refproof}{thm:bairetopolish} \begin{refproof}{thm:bairetopolish}
Let $d$ be a complete metric on $X$. Let $d$ be a complete metric on $X$.
W.l.o.g.~$\diam(X) \le 1$. W.l.o.g.~$\diam(X) \le 1$.
@ -220,7 +230,7 @@
\item $F_\emptyset = X$, \item $F_\emptyset = X$,
\item $F_s$ is $F_\sigma$ for all $s$. \item $F_s$ is $F_\sigma$ for all $s$.
\item The $F_{s \concat i}$ partition $F_s$, \item The $F_{s \concat i}$ partition $F_s$,
i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$. % TODO change notation? i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.
Furthermore we want that Furthermore we want that
$\overline{F_{s \concat i}} \subseteq F_s$ $\overline{F_{s \concat i}} \subseteq F_s$
@ -228,6 +238,7 @@
\item $\diam(F_s) \le 2^{-|s|}$. \item $\diam(F_s) \le 2^{-|s|}$.
\end{enumerate} \end{enumerate}
\gist{%
Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$. Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
We need to construct a partition $(F_i)_{i \in \N}$ We need to construct a partition $(F_i)_{i \in \N}$
of $F$ with $\overline{F_i} \subseteq F$ of $F$ with $\overline{F_i} \subseteq F$
@ -252,6 +263,7 @@
The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$ The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
are $F_\sigma$, disjoint are $F_\sigma$, disjoint
and $F_i^0 = \bigcup_{j} D_j$. and $F_i^0 = \bigcup_{j} D_j$.
}{Induction.}

74
inputs/lecture_04.tex
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@ -5,6 +5,7 @@
\end{remark} \end{remark}
\begin{refproof}{thm:bairetopolish} \begin{refproof}{thm:bairetopolish}
\gist{%
Take Take
\[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\] \[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
@ -12,16 +13,19 @@
we have we have
\[ \[
\bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}. \bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
\] \]
}{}
$f\colon D \to X$ is determined by $f\colon D \to X$ is determined by
\[ \[
\{f(x)\} = \bigcap_{n} F_{x\defon{n}} \{f(x)\} = \bigcap_{n} F_{x\defon{n}}
\] \]
$f$ is injective and continuous. \gist{%
The proof of this is exactly the same as in $f$ is injective and continuous.
\yaref{thm:cantortopolish}. The proof of this is exactly the same as in
\yaref{thm:cantortopolish}.
}{}
\begin{claim} \begin{claim}
\label{thm:bairetopolish:c1} \label{thm:bairetopolish:c1}
@ -47,7 +51,7 @@
there exists $y = \lim_n f(x_n)$. there exists $y = \lim_n f(x_n)$.
Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$, Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
we get that $y \in \overline{F_{x\defon{N}}}$. we get that $y \in \overline{F_{x\defon{N}}}$.
Note that for $N' > N$ by the same argument Note that for $N' > N$ by the same argument
we get $y \in \overline{F_{x\defon{N'}}}$. we get $y \in \overline{F_{x\defon{N'}}}$.
Hence Hence
@ -55,20 +59,20 @@
i.e.~$y \in D$ and $y = f(x)$. i.e.~$y \in D$ and $y = f(x)$.
\end{refproof} \end{refproof}
We extend $f$ to $g\colon\cN \to X$ We extend $f$ to $g\colon\cN \to X$
in the following way: in the following way:
Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$. Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
Clearly $S$ is a pruned tree. Clearly $S$ is a pruned tree.
Moreover, since $D$ is closed, we have that\todo{Proof this (homework?)} Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1})
\[ \[
D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}. D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
\] \]
We construct a \vocab{retraction} $r\colon\cN \to D$ We construct a \vocab{retraction} $r\colon\cN \to D$
(i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection). (i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).
Then $g \coloneqq f \circ r$. Then $g \coloneqq f \circ r$.
To construct $r$, we will define To construct $r$, we will define
$\phi\colon \N^{<\N} \to S$ by induction on the length $\phi\colon \N^{<\N} \to S$ by induction on the length
such that such that
\begin{itemize} \begin{itemize}
@ -76,21 +80,27 @@
\item $|s| = \phi(|s|)$, \item $|s| = \phi(|s|)$,
\item if $s \in S$, then $\phi(s) = s$. \item if $s \in S$, then $\phi(s) = s$.
\end{itemize} \end{itemize}
Let $\phi(\emptyset) = \emptyset$. \gist{%
Suppose that $\phi(t)$ is defined. Let $\phi(\emptyset) = \emptyset$.
If $t\concat a \in S$, then set Suppose that $\phi(t)$ is defined.
$\phi(t\concat a) \coloneqq t\concat a$. If $t\concat a \in S$, then set
Otherwise take some $b$ such that $\phi(t\concat a) \coloneqq t\concat a$.
$t\concat b \in S$ and define Otherwise take some $b$ such that
$\phi(t\concat a) \coloneqq \phi(t)\concat b$. $t\concat b \in S$ and define
$\phi(t\concat a) \coloneqq \phi(t)\concat b$.%
}{}%
This is possible since $S$ is pruned. This is possible since $S$ is pruned.
Let $r\colon \cN = [\N^{<\N}] \to [S] = D$ \gist{%
be the function defined by $r(x) = \bigcup_n f(x\defon{n})$. Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
}{}
$r$ is continuous, since $r$ is continuous, since
$d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz $d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz
It is immediate that $r$ is a retraction. \gist{%
It is immediate that $r$ is a retraction.
}{}
\end{refproof} \end{refproof}
\section{Meager and Comeager Sets} \section{Meager and Comeager Sets}
@ -110,42 +120,48 @@
then $A \cap U$ is not dense in $U$). then $A \cap U$ is not dense in $U$).
\end{itemize} \end{itemize}
A set $B \subseteq X$ is \vocab{meager} A set $B \subseteq X$ is \vocab{meager}
(or \vocab{first category}), (or \vocab{first category}),
iff it is a countable union of nwd sets. iff it is a countable union of nwd sets.
The complement of a meager set is called The complement of a meager set is called
\vocab{comeager}. \vocab{comeager}.
\end{definition} \end{definition}
\gist{%
\begin{example} \begin{example}
$\Q \subseteq \R$ is meager. $\Q \subseteq \R$ is meager.
\end{example} \end{example}
}{}
\begin{notation} \begin{notation}
Let $A, B \subseteq X$. Let $A, B \subseteq X$.
We write $A =^\ast B$ We write $A =^\ast B$
iff the \vocab{symmetric difference}, iff the \vocab{symmetric difference},
$A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$, $A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$,
is meager. is meager.
\end{notation} \end{notation}
\gist{%
\begin{remark} \begin{remark}
$=^\ast$ is an equivalence relation. $=^\ast$ is an equivalence relation.
\end{remark} \end{remark}
}{}
\begin{definition} \begin{definition}
A set $A \subseteq X$ A set $A \subseteq X$
has the \vocab{Baire property} (\vocab{BP}) has the \vocab{Baire property} (\vocab{BP})
if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$. if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
\end{definition} \end{definition}
\gist{%
Note that open sets and meager sets have the Baire property. Note that open sets and meager sets have the Baire property.
}{}
\gist{%
\begin{example} \begin{example}
\begin{itemize} \begin{itemize}
\item $\Q \subseteq \R$ is $F_\sigma$. \item $\Q \subseteq \R$ is $F_\sigma$.
\item $\R \setminus \Q \subseteq \R$ is $G_\delta$. \item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
\item $\Q \subseteq \R$ is not $G_{\delta}$. \item $\Q \subseteq \R$ is not $G_{\delta}$:
(It is dense and meager, It is dense and meager,
hence it can not be $G_\delta$ hence it can not be $G_\delta$
by the Baire category theorem). by the \yaref{thm:bct}.
\end{itemize} \end{itemize}
\end{example} \end{example}
}{}

30
inputs/lecture_05.tex
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@ -6,10 +6,9 @@
\item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense. \item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense.
\item Any meager set $B$ is contained in a meager $F_{\sigma}$-set. \item Any meager set $B$ is contained in a meager $F_{\sigma}$-set.
\end{itemize} \end{itemize}
\end{fact} \end{fact}
\begin{proof} % remove? \gist{%
\begin{proof}
\begin{itemize} \begin{itemize}
\item This follows from the definition as $\overline{\overline{A}} = \overline{A}$. \item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.
\item Trivial. \item Trivial.
@ -17,7 +16,9 @@
Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$. Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.
\end{itemize} \end{itemize}
\end{proof} \end{proof}
}{}
\gist{%
\begin{definition} \begin{definition}
A \vocab{$\sigma$-algebra} on a set $X$ A \vocab{$\sigma$-algebra} on a set $X$
is a collection of subsets of $X$ is a collection of subsets of $X$
@ -32,14 +33,15 @@
Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$ Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$
we have that $\sigma$-algebras are closed under countable intersections. we have that $\sigma$-algebras are closed under countable intersections.
\end{fact} \end{fact}
}{}
\begin{theorem} \begin{theorem}
\label{thm:bairesigma} \label{thm:bairesigma}
Let $X$ be a topological space. Let $X$ be a topological space.
Then the collection of sets with the Baire property Then the collection of sets with the Baire property
is a $\sigma$-algebra on $X$. is \gist{a $\sigma$-algebra on $X$.
It is the smallest $\sigma$-algebra It is}{} the smallest $\sigma$-algebra
containing all meager and open sets. containing all meager and open sets.
\end{theorem} \end{theorem}
\begin{refproof}{thm:bairesigma} \begin{refproof}{thm:bairesigma}
@ -93,10 +95,11 @@
% Nwd set of positive measure. % Nwd set of positive measure.
% TODO % TODO
% remove open intervals such that their length does not add to 0 % remove open intervals such that their length does not add to 0
% %
%\end{example} %\end{example}
\begin{theorem}[Baire Category theorem] \begin{theorem}[Baire Category theorem]
\yalabel{Baire Category Theorem}{Baire Category Theorem}{thm:bct}
Let $X$ be a completely metrizable space. Let $X$ be a completely metrizable space.
Then every comeager set of $X$ is dense in $X$. Then every comeager set of $X$ is dense in $X$.
\end{theorem} \end{theorem}
@ -111,8 +114,8 @@
\item The intersection of countable many \item The intersection of countable many
open dense sets is dense. open dense sets is dense.
\end{enumerate} \end{enumerate}
In this case $X$ is called a \vocab{Baire space}. In this case $X$ is called a \vocab{Baire space}.%
\footnote{see \yaref{s5e1}} \footnote{cf.~\yaref{s5e1}}
\end{theoremdef} \end{theoremdef}
\begin{proof} \begin{proof}
\todo{Proof (short)} \todo{Proof (short)}
@ -124,12 +127,11 @@
We have that We have that
\[ \[
\emptyset = \bigcap_{n} (X \setminus \overline{A_n}) \emptyset = \bigcap_{n} (X \setminus \overline{A_n})
\] \]
is dense by (iii). is dense by (iii).
This proof can be adapted to other open sets $X$. This proof can be adapted to other open sets $X$.
\end{proof} \end{proof}
\begin{notation} \begin{notation}
Let $X ,Y$ be topological spaces, Let $X ,Y$ be topological spaces,
$A \subseteq X \times Y$ $A \subseteq X \times Y$
@ -138,12 +140,12 @@
Let Let
\[ \[
A_x \coloneqq \{y \in Y : (x,y) \in A\} A_x \coloneqq \{y \in Y : (x,y) \in A\}
\] \]
and and
\[ \[
A^y \coloneqq \{x \in X : (x,y) \in A\} . A^y \coloneqq \{x \in X : (x,y) \in A\} .
\] \]
\end{notation} \end{notation}
The following similar to Fubini, The following similar to Fubini,
@ -273,9 +275,11 @@ but for meager sets:
% \end{refproof} % \end{refproof}
% TODO fix claim numbers % TODO fix claim numbers
\gist{%
\begin{remark} \begin{remark}
Suppose that $A$ has the BP. Suppose that $A$ has the BP.
Then there is an open $U$ such that Then there is an open $U$ such that
$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager. $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
Then $A = U \symdif M$. Then $A = U \symdif M$.
\end{remark} \end{remark}
}{}

7
inputs/lecture_06.tex
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@ -98,14 +98,13 @@ Let $X$ be a topological space.
Then define Then define
\[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\] \[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\]
\[ \[
\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq \Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq
\{X \setminus A | A \in \Sigma^0_\alpha(X)\}, \{X \setminus A | A \in \Sigma^0_\alpha(X)\},
\] \]
% \todo{Define $\lnot$ (element-wise complement)}
and for $\alpha > 1$ and for $\alpha > 1$
\[ \[
\Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}. \Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.
\] \]
Note that $\Pi_1^0$ is the set of closed sets, Note that $\Pi_1^0$ is the set of closed sets,
$\Sigma^0_2 = F_\sigma$, $\Sigma^0_2 = F_\sigma$,

78
inputs/lecture_07.tex
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@ -16,17 +16,17 @@
We have that We have that
\[ \[
\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}. \Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
\] \]
Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$. Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.
We have We have
\[ \[
\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X). \cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).
\] \]
Hence Hence
\[ \[
|\cB(X)| \le \omega_1 \cdot \fc = \fc. |\cB(X)| \le \omega_1 \cdot \fc = \fc.
\] \]
\end{proof} \end{proof}
\begin{proposition}[Closure properties] \begin{proposition}[Closure properties]
@ -37,54 +37,58 @@
\item \begin{itemize} \item \begin{itemize}
\item $\Sigma^0_\xi(X)$ is closed under countable unions. \item $\Sigma^0_\xi(X)$ is closed under countable unions.
\item $\Pi^0_\xi(X)$ is closed under countable intersections. \item $\Pi^0_\xi(X)$ is closed under countable intersections.
\item $\Delta^0_\xi(X)$ is closed under complements, \item $\Delta^0_\xi(X)$ is closed under complements.
countable unions and
countable intersections.
\end{itemize} \end{itemize}
\item \begin{itemize} \item \begin{itemize}
\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections. \item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
\item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions. \item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
\item $\Delta^0_\xi(X)$ is closed under finite unions and
finite intersections.
\end{itemize} \end{itemize}
\end{enumerate} \end{enumerate}
\end{proposition} \end{proposition}
\gist{%
\begin{proof} \begin{proof}
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item This follows directly from the definition. \item This follows directly from the definition.
Note that a countable intersection can be written Note that a countable intersection can be written
as a complement of the countable union of complements: as a complement of the countable union of complements:
\[ \[
\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}. \bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
\] \]
\item If suffices to check this for $\Sigma^0_{\xi}(X)$. \item If suffices to check this for $\Sigma^0_{\xi}(X)$.
Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$ Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$. and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
Then Then
\[ \[
A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right) A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)
\] \]
and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$. and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
\end{enumerate} \end{enumerate}
\end{proof} \end{proof}
}{}
\begin{example} \begin{example}
Consider the cantor space $2^{\omega}$. Consider the cantor space $2^{\omega}$.
We have that $\Delta^0_1(2^{\omega})$ We have that $\Delta^0_1(2^{\omega})$
is not closed under countable unions is not closed under countable unions%
(countable unions yield all open sets, but there are open \gist{ (countable unions yield all open sets, but there are open
sets that are not clopen). sets that are not clopen)}{}.
\end{example} \end{example}
\subsection{Turning Borels Sets into Clopens} \subsection{Turning Borels Sets into Clopens}
\begin{theorem}% \begin{theorem}%
\gist{%
\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}'' \footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
unfortunately seems to be non-standard vocabulary. unfortunately seems to be non-standard vocabulary.
Our tutor repeatedly advised against using it in the final exam. Our tutor repeatedly advised against using it in the final exam.
Contrary to popular belief Contrary to popular belief
the very same tutor was \textit{not} the one first to introduce it, the very same tutor was \textit{not} the one first to introduce it,
as it would certainly be spelled ``to clopenise'' if that were the case. as it would certainly be spelled ``to clopenise'' if that were the case.
} }%
}{}%
\label{thm:clopenize} \label{thm:clopenize}
Let $(X, \cT)$ be a Polish space. Let $(X, \cT)$ be a Polish space.
For any Borel set $A \subseteq X$, For any Borel set $A \subseteq X$,
@ -163,7 +167,7 @@
such that $\cT_n \supseteq \cT$ such that $\cT_n \supseteq \cT$
and $\cB(\cT_n) = \cB(\cT)$. and $\cB(\cT_n) = \cB(\cT)$.
Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$ Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
is still Polish is Polish
and $\cB(\cT_\infty) = \cB(T)$. and $\cB(\cT_\infty) = \cB(T)$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
@ -183,7 +187,51 @@
definition of $\cF$ belong to definition of $\cF$ belong to
a countable basis of the respective $\cT_n$). a countable basis of the respective $\cT_n$).
\todo{This proof will be finished in the next lecture} % Proof was finished in lecture 8
Let $Y = \prod_{n \in \N} (X, \cT_n)$.
Then $Y$ is Polish.
Let $\delta\colon (X, \cT_\infty) \to Y$
defined by $\delta(x) = (x,x,x,\ldots)$.
\begin{claim}
$\delta$ is a homeomorphism.
\end{claim}
\begin{subproof}
Clearly $\delta$ is a bijection.
We need to show that it is continuous and open.
Let $U \in \cT_i$.
Then
\[
\delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
\]
hence $\delta$ is continuous.
Let $U \in \cT_\infty$.
Then $U$ is the union of sets of the form
\[
V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
\]
for some $n_1 < n_2 < \ldots < n_u$
and $U_{n_i} \in \cT_i$.
Thus is suffices to consider sets of this form.
We have that
\[
\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.
\]
\end{subproof}
This will finish the proof since
\[
D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
\]
Why? Let $(x_n) \in Y \setminus D$.
Then there are $i < j$ such that $x_i \neq x_j$.
Take disjoint open $x_i \in U$, $x_j \in V$.
Then
\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]
is open in $Y\setminus D$.
Hence $Y \setminus D$ is open, thus $D$ is closed.
It follows that $D$ is Polish.
\end{proof} \end{proof}
We need to show that $A$ is closed under countable unions. We need to show that $A$ is closed under countable unions.

63
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@ -1,61 +1,8 @@
\lecture{08}{2023-11-10}{} \lecture{08}{2023-11-10}{}\footnote{%
In the beginning of the lecture, we finished
\todo{put this lemma in the right place} the proof of \yaref{thm:clopenize:l2}.
\begin{lemma}[Lemma 2] This has been moved to the notes on lecture 7.%
Let $(X, \cT)$ be a Polish space. }
Let $\cT_n \supseteq \cT$ be Polish
with $\cB(X, \cT_n) = \cB(X, \cT)$.
Let $\cT_\infty$ be the topology generated
by $\bigcup_n \cT_n$.
Then $(X, \cT_\infty)$ is Polish
and $\cB(X, \cT_\infty) = \cB(X, \cT)$.
\end{lemma}
\begin{proof}
Let $Y = \prod_{n \in \N} (X, \cT_n)$.
Then $Y$ is Polish.
Let $\delta\colon (X, \cT_\infty) \to Y$
defined by $\delta(x) = (x,x,x,\ldots)$.
\begin{claim}
$\delta$ is a homeomorphism.
\end{claim}
\begin{subproof}
Clearly $\delta$ is a bijection.
We need to show that it is continuous and open.
Let $U \in \cT_i$.
Then
\[
\delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
\]
hence $\delta$ is continuous.
Let $U \in \cT_\infty$.
Then $U$ is the union of sets of the form
\[
V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
\]
for some $n_1 < n_2 < \ldots < n_u$
and $U_{n_i} \in \cT_i$.
Thus is suffices to consider sets of this form.
We have that
\[
\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.
\]
\end{subproof}
This will finish the proof since
\[
D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
\]
Why? Let $(x_n) \in Y \setminus D$.
Then there are $i < j$ such that $x_i \neq x_j$.
Take disjoint open $x_i \in U$, $x_j \in V$.
Then
\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]
is open in $Y\setminus D$.
Hence $Y \setminus D$ is open, thus $D$ is closed.
It follows that $D$ is Polish.
\end{proof}
\subsection{Parametrizations} \subsection{Parametrizations}
%\todo{choose better title} %\todo{choose better title}

1
inputs/lecture_09.tex
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@ -148,6 +148,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
\begin{theorem} \begin{theorem}
\label{thm:universals11}
Let $X,Y$ be uncountable Polish spaces. Let $X,Y$ be uncountable Polish spaces.
There exists a $Y$-universal $\Sigma^1_1(X)$ set. There exists a $Y$-universal $\Sigma^1_1(X)$ set.
\end{theorem} \end{theorem}

20
inputs/lecture_11.tex
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@ -153,13 +153,14 @@ We will not proof this in this lecture.
\subsection{Ill-Founded Trees} \subsection{Ill-Founded Trees}
\gist{%
Recall that a \vocab{tree} on $\N$ is a subset of Recall that a \vocab{tree} on $\N$ is a subset of
$\N^{<\N}$ $\N^{<\N}$
closed under taking initial segments. closed under taking initial segments.
We now identify trees with their characteristic functions, We now identify trees with their characteristic functions,
i.e.~we want to associate a tree $T \subseteq \N^{<\N}$ i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}%
{We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:}
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\ \One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\
x &\longmapsto & \begin{cases} x &\longmapsto & \begin{cases}
@ -167,16 +168,15 @@ i.e.~we want to associate a tree $T \subseteq \N^{<\N}$
0 &: x \not\in T. 0 &: x \not\in T.
\end{cases} \end{cases}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Note that $\One_T \in {\{0,1\}^\N}^{< \N}$. \gist{Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.}{}
Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$. Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
\begin{observe} \begin{observe}
\[ $\Tr \subseteq {2^{\N}}^{<\N}$ is closed
\Tr \subseteq {2^{\N}}^{<\N} (where we take the topology of the Cantor space).
\]
is closed (where we take the topology of the Cantor space).
\end{observe} \end{observe}
\gist{%
Indeed, for any $ s \in \N^{<\N}$ Indeed, for any $ s \in \N^{<\N}$
we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$ we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen. and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
@ -185,6 +185,4 @@ In particular for $s$ fixed,
we have that we have that
\[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\] \[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
is clopen in ${2^{\N}}^{<\N}$. is clopen in ${2^{\N}}^{<\N}$.
}{}

239
inputs/lecture_12.tex
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@ -21,102 +21,135 @@
T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1. T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1.
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Consider $D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$. Consider
Note that this set is closed in $\Tr \times \cN$, \[D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
since it is a countable intersection of clopen sets. \gist{%
% TODO Why clopen? Note that this set is closed in $\Tr \times \cN$,
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$. since it is a countable intersection of clopen sets.
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
}{$D \overset{\text{closed}}{\subseteq} \Tr \times \cN$ and $\IF = \proj_{\Tr}(D)$.}
\end{proof} \end{proof}
\begin{definition} \begin{definition}
An analytic set $B$ in some Polish space $Y$ An analytic set $B$ in some Polish space $Y$
is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete}) is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete})
\gist{%
iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$, iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$,
there exists a Borel function $f\colon X\to Y$ there exists a Borel function $f\colon X\to Y$
such that $x \in A \iff f(x) \in B$. such that $x \in A \iff f(x) \in B$,
i.e.~$f^{-1}(B) = A$.
}{%
iff for any $A \in \Sigma^1_1(X)$, $X$ Polish
there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.}
Similarly, define \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}). \gist{%
Similarly, a conalytic set $B$ is called
\vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete})
iff for any $A \subseteq \Pi^1_1(X)$
there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.
}{Similarly we define \vocab{complete coanalytic} / \vocab{$\Pi_1^1$-complete}.}
\end{definition} \end{definition}
\begin{observe} \begin{observe}
\leavevmode \leavevmode
\begin{itemize} \begin{itemize}
\item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete. \item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete.
\item $\Sigma^1_1$-complete sets are never Borel: \item $\Sigma^1_1$-complete sets are never Borel%
Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$. \gist{:
Take $A \in \Sigma^1_1(X) \setminus \cB(X)$ Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
and $f\colon X \to Y$ Borel. Take $A \in \Sigma^1_1(X) \setminus \cB(X)$%
But then $f^{-1}(B)$ is Borel. \footnote{e.g.~\yaref{thm:universals11}}
and $f\colon X \to Y$ Borel.
But then we get that $f^{-1}(B)$ is Borel $\lightning$.
}{.}
\end{itemize} \end{itemize}
\end{observe} \end{observe}
\begin{theorem} \begin{theorem}
\label{thm:lec12:1} \label{thm:lec12:1}
Suppose that $A \subseteq \cN$ is analytic. Suppose that $A \subseteq \cN$ is analytic.
Then there is $f\colon \cN \to \Tr$\todo{Borel?} \gist{%
such that $x \in A \iff f(x)$ is ill-founded. Then there is a continuous function $f\colon \cN \to \Tr$
such that $x \in A \iff f(x)$ is ill-founded,
i.e.~$A = f^{-1}(\IF)$.
}{%
Then there exists $f\colon \cN \to \Tr$ continuous
such that $A = f^{-1}(\IF)$.
}
\end{theorem} \end{theorem}
For the proof we need some prerequisites: For the proof we need some prerequisites:
\begin{enumerate}[1.]
\item Recall that for $S$ countable, \gist{%
the pruned\footnote{no maximal elements, in particular this implies ill-founded if the tree is non empty.} trees Recall that for $S$ countable,
$T \subseteq S^{<\N}$ on $S$ correspond the pruned%
to closed subsets of $S^{\N}$: \footnote{no maximal elements,
\begin{IEEEeqnarray*}{rCl} in particular this implies ill-founded if the tree is non empty.
T &\longmapsto & [T]\\ } trees $T \subseteq S^{<\N}$ on $S$ correspond
\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\ to closed subsets of $S^{\N}$:%
\end{IEEEeqnarray*} \footnote{cf.~\yaref{s3e1} (c)}
\todo{Copy from exercises} \begin{IEEEeqnarray*}{rCl}
\item \leavevmode\begin{definition} T &\longmapsto & [T]\\
If $T$ is a tree on $\N \times \N$ \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
and $x \in \cN$, \end{IEEEeqnarray*}
then the \vocab{section at $x$} }{%
%denoted $T(x)$, For $S$ countable,
is the following tree on $\N$ : pruned trees on $S$ correspond to closed subsets of $S^{\N}$
\[ via $T \mapsto [T]$.
T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}. }
\] \begin{definition}
\end{definition} If $T$ is a tree on $\N \times \N$
\item \leavevmode and $x \in \cN$,
\begin{proposition} then the \vocab{section at $x$}
\label{prop:lec12:2} denoted $T(x)$,
Let $A \subseteq \cN$. is the following tree on $\N$ :
The following are equivalent: \[
\begin{itemize} T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
\item $A$ is analytic. \]
\item There is a pruned tree on $\N \times \N$ \end{definition}
such that \begin{proposition}
\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\] \label{prop:lec12:2}
\end{itemize} Let $A \subseteq \cN$.
\end{proposition} The following are equivalent:
\begin{proof} \begin{itemize}
$A$ is analytic iff \item $A$ is analytic.
there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$ \item There is a pruned tree on $\N \times \N$
such that $A = \proj_1(F)$. such that
But closed sets of $\N \times \N$ correspond to pruned trees, \[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
by the first point. \end{itemize}
\end{proof} \end{proposition}
\end{enumerate} \begin{proof}
\gist{%
$A$ is analytic iff
there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$
such that $A = \proj_1(F)$.
But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees,
by the first point.
}{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.}
\end{proof}
\begin{refproof}{thm:lec12:1} \begin{refproof}{thm:lec12:1}
Take a tree $T$ on $\N \times \N$ \gist{%
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$. Take a tree $T$ on $\N \times \N$
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
}{Write $A = \proj_1([T])$ for a pruned tree $T$ on $\N \times \N$.}
Consider Consider
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
f\colon \cN &\longrightarrow & \Tr \\ f\colon \cN &\longrightarrow & \Tr \\
x &\longmapsto & T(x). x &\longmapsto & T(x).
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Clearly $x \in A \iff f(x)$ is ill-founded. \gist{%
$f$ is continuous: Clearly $x \in A \iff f(x) \in \IF$.
Let $x\defon{n} = y\defon{n}$ for some $n \in \N$. $f$ is continuous:
Then for all $m \le n, s,t \in \N^{<\N}$ Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$, Then for all $m \le n, s,t \in \N^{<\N}$
we have such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
\begin{itemize} we have
\item $t \in T(x) \iff (s,t) \in T$, \begin{itemize}
\item $t \in T(y) \iff (s,t) \in T$. \item $t \in T(x) \iff (s,t) \in T$,
\end{itemize} \item $t \in T(y) \iff (s,t) \in T$.
\end{itemize}
So if $x\defon{n} = y\defon{n}$, So if $x\defon{n} = y\defon{n}$,
then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.. then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.
}{}
\end{refproof} \end{refproof}
\begin{corollary} \begin{corollary}
@ -125,7 +158,9 @@ For the proof we need some prerequisites:
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Let $X$ be Polish. Let $X$ be Polish.
Suppose that $A \subseteq X$ is analytic and uncountable. Suppose that $A \subseteq X$ is analytic and uncountable%
\gist{}{ (trivial for countable)}.
Then Then
% https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ== % https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==
\[\begin{tikzcd} \[\begin{tikzcd}
@ -138,16 +173,18 @@ For the proof we need some prerequisites:
\end{tikzcd}\] \end{tikzcd}\]
where $f$ is chosen as in \yaref{thm:lec12:1}. where $f$ is chosen as in \yaref{thm:lec12:1}.
If $X$ is Polish and countable and $A \subseteq X$ analytic, \gist{%
just consider If $X$ is Polish and countable and $A \subseteq X$ analytic,
\begin{IEEEeqnarray*}{rCl} just consider
g \colon X &\longrightarrow & \Tr \\ \begin{IEEEeqnarray*}{rCl}
x &\longmapsto & \begin{cases} g \colon X &\longrightarrow & \Tr \\
a &: x \in A,\\ x &\longmapsto & \begin{cases}
b &: x \not\in A,\\ a &: x \in A,\\
\end{cases} b &: x \not\in A,\\
\end{IEEEeqnarray*} \end{cases}
where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily. \end{IEEEeqnarray*}
where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
}{}
\end{proof} \end{proof}
\subsection{Linear Orders} \subsection{Linear Orders}
@ -161,18 +198,19 @@ Let
\[ \[
\WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}. \WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}.
\] \]
\gist{%
Recall that
\begin{itemize}
\item $(A,<)$ is a well ordering iff there are no infinite descending chains.
\item Every well ordering is isomorphic to an ordinal.
\item Any two well orderings are comparable,
i.e.~they are isomorphic,
or one is isomorphic to an initial segment of the other.
Recall that Let $(A, <_A) \prec (B, <_B)$ denote that
\begin{itemize} $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
\item $(A,<)$ is a well ordering iff there are no infinite descending chains. \end{itemize}
\item Every well ordering is isomorphic to an ordinal. }{}
\item Any two well orderings are comparable,
i.e.~they are isomorphic,
or one is isomorphic to an initial segment of the other.
Let $(A, <_A) \prec (B, <_B)$ denote that
$(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
\end{itemize}
\begin{definition} \begin{definition}
A \vocab{rank} on some set $C$ A \vocab{rank} on some set $C$
@ -181,13 +219,14 @@ Recall that
\phi\colon C \to \Ord. \phi\colon C \to \Ord.
\] \]
\end{definition} \end{definition}
\begin{example} \gist{%
Let $C = \WO$ \begin{example}
and Let $C = \WO$
\begin{IEEEeqnarray*}{rCl} and
\phi\colon \WO &\longrightarrow & \Ord \\ \begin{IEEEeqnarray*}{rCl}
\end{IEEEeqnarray*} \phi\colon \WO &\longrightarrow & \Ord \\
where $\phi((A,<_A))$ is the unique ordinal \end{IEEEeqnarray*}
isomorphic to $(A, <_A)$. where $\phi((A,<_A))$ is the unique ordinal
\end{example} isomorphic to $(A, <_A)$.
\end{example}
}{}

43
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@ -1,11 +1,12 @@
\lecture{13}{2023-11-08}{} \lecture{13}{2023-11-08}{}
\gist{%
% Recap % Recap
$\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $. $\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
$\LO \subseteq 2^{\N \times \N}$ is closed $\LO \subseteq 2^{\N \times \N}$ is closed
and $\WO = \{x \in \LO: x \text{ is a wellordering}\} $ and $\WO = \{x \in \LO: x \text{ is a wellordering}\} $
is coanalytic in $\LO$. is coanalytic in $\LO$.
% End Recap % End Recap
}{}
Another way to code linear orders: Another way to code linear orders:
@ -32,6 +33,7 @@ with $(f^{-1}(\{1\}), <)$.
and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$. and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
\end{proof} \end{proof}
\begin{definition}[\vocab{Kleene-Brouwer ordering}] \begin{definition}[\vocab{Kleene-Brouwer ordering}]
Let $(A,<)$ be a linear order and $A$ countable. Let $(A,<)$ be a linear order and $A$ countable.
We define the linear order $<_{KB}$ on $A^{<\N}$ We define the linear order $<_{KB}$ on $A^{<\N}$
@ -55,25 +57,30 @@ with $(f^{-1}(\{1\}), <)$.
$(T, <_{KB}\defon{T})$ is well ordered. $(T, <_{KB}\defon{T})$ is well ordered.
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
If $T$ is ill-founded and $x \in [T]$, \gist{%
then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$. If $T$ is ill-founded and $x \in [T]$,
Thus $(T, <_{KB}\defon{T})$ is not well ordered. then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
Thus $(T, <_{KB}\defon{T})$ is not well ordered.
Conversely, let $<\defon{KB}$ be not a well-ordering Conversely, let $<\defon{KB}$ be not a well-ordering
on $T$. on $T$.
Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$ Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
be an infinite descending chain. be an infinite descending chain.
We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$ We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
stabilizes for $n > n_0$. stabilizes for $n > n_0$.
Let $a_0 \coloneqq s_{n_0}(0)$. Let $a_0 \coloneqq s_{n_0}(0)$.
Now for $n \ge n_0$ we have that $s_n(0)$ is constant, Now for $n \ge n_0$ we have that $s_n(0)$ is constant,
hence for $n > n_0$ the value $s_{n}(1)$ must be defined. hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
Thus there is $n_1 \ge n_0$ such that $s_n(1)$ Thus there is $n_1 \ge n_0$ such that $s_n(1)$
is constant for all $n \ge n_1$. is constant for all $n \ge n_1$.
Let $a_1 \coloneqq s_{n_1}(1)$ Let $a_1 \coloneqq s_{n_1}(1)$
and so on. and so on.
Then $(a_0,a_1,a_2, \ldots) \in [T]$. Then $(a_0,a_1,a_2, \ldots) \in [T]$.
}{easy}
\end{proof} \end{proof}
% TODO ANKI-MARKER
\begin{theorem}[Lusin-Sierpinski] \begin{theorem}[Lusin-Sierpinski]
The set $\LO \setminus \WO$ The set $\LO \setminus \WO$
(resp.~$2^{\Q} \setminus \WO$) (resp.~$2^{\Q} \setminus \WO$)

9
inputs/lecture_14.tex
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@ -66,7 +66,7 @@
} }
in $(y, <_\Q)$ is in $(y, <_\Q)$ is
cofinal in $(y, <_{\Q})$ and vice versa. cofinal in $(y, <_{\Q})$ and vice versa.
Equivalently, either $(x <^\ast_\phi y)$ Equivalently, either $(x <^\ast_\phi y)$
or or
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
& &x,y \in \WO\\ & &x,y \in \WO\\
@ -84,10 +84,9 @@
such that such that
\[ \[
\forall x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast). \forall x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast).
\] \]
We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$. We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.%
\todo{missing picture \footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}}
\url{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}}
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}

11
inputs/lecture_16.tex
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@ -175,9 +175,10 @@ By Zorn's lemma, this will follow from
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
The topology induced by the metric The topology induced by the metric
is given by basic open subsets\footnote{see Exercise Sheet 9% TODO REF is given by basic open subsets\footnote{cf.~\yaref{s9e2}}
} of the form
$[U_0; U_1,\ldots, U_n]$, $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$, $[U_0; U_1,\ldots, U_n]$,
for $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
where where
\[ \[
[U_0; U_1,\ldots,U_n] \coloneqq [U_0; U_1,\ldots,U_n] \coloneqq
@ -188,8 +189,8 @@ By Zorn's lemma, this will follow from
We want to view flows as a metric space. We want to view flows as a metric space.
For a fixed compact metric space $X$, For a fixed compact metric space $X$,
we can view the flows $(X,\Z)$ as a subset of $\cC(X,X)$. we can view the flows $(X,\Z)$ as a subset of $\cC(X,X)$.
Note that $\cC(X,X)$ is Polish. Note that $\cC(X,X)$ is Polish.\footnote{cf.~\yaref{s1e4}}
Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$. Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.\footnote{Exercise} % TODO
However we do not want to consider only flows on a fixed space $X$, However we do not want to consider only flows on a fixed space $X$,
but we want to look all flows at the same time. but we want to look all flows at the same time.

13
inputs/tutorial_01.tex
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@ -27,8 +27,10 @@
\begin{definition} \begin{definition}
A topological space is \vocab{Lindelöf} A topological space is \vocab{Lindelöf}
if every open cover has a countable subcover. iff every open cover has a countable subcover.
\end{definition} \end{definition}
\begin{fact} \begin{fact}
Let $X$ be a metric space. Let $X$ be a metric space.
If $X$ is Lindelöf, If $X$ is Lindelöf,
@ -64,5 +66,12 @@
and Lindelöf coincide. and Lindelöf coincide.
In arbitrary topological spaces, In arbitrary topological spaces,
Lindelöf is the strongest of these notions. Lindelöf is the weakest of these notions.
\end{remark} \end{remark}
\begin{definition}+
A metric space $X$ is \vocab{totally bounded}
iff for every $\epsilon > 0$ there exists
a finite set of points $x_1,\ldots,x_n$
such that $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$.
\end{definition}

18
inputs/tutorial_09.tex
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@ -140,16 +140,7 @@ amounts to a finite number of conditions on the preimage.
\end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X). \end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X).
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Clearly this has the desired properties. Clearly this has the desired properties.
\item We have \item Let $X$ be a compact Polish space.
\begin{IEEEeqnarray*}{Cl}
& \Z \circlearrowright X \text{ has a dense orbit}\\
\iff& \exists x \in X.~ \overline{\Z\cdot x} = X\\
\iff& \exists x \in X.~\forall U\overset{\text{open}}{\subseteq} X.~\exists z \in \Z.~
z \cdot x \in U\\
\iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~
\exists z \in \Z.~f^z(x) \in U.
\end{IEEEeqnarray*}
\item Let $X$ be a compact Polish space.
What is the Borel complexity of $\Homeo(X)$ inside $\cC(X,X)$? What is the Borel complexity of $\Homeo(X)$ inside $\cC(X,X)$?
Recall that $\cC(X,X)$ is a Polish space with the uniform topology. Recall that $\cC(X,X)$ is a Polish space with the uniform topology.
@ -160,14 +151,9 @@ amounts to a finite number of conditions on the preimage.
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
by the general fact by the general fact
\begin{fact} \begin{fact}
Let $X$ be comapct and $Y$ Hausdorff, Let $X$ be compact and $Y$ Hausdorff,
$f\colon X \to Y$ a continuous bijection. $f\colon X \to Y$ a continuous bijection.
Then $f$ is a homeomorphism. Then $f$ is a homeomorphism.
\end{fact} \end{fact}
\item It suffices to check the condition from part (b)
for open sets $U$ of a countable basis
and points $x \in X$ belonging to a countable dense subset.
Replacing quantifiers by unions resp.~intersections
gives that $D$ is Borel.
\end{itemize} \end{itemize}

146
inputs/tutorial_13.tex Normal file
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@ -0,0 +1,146 @@
\tutorial{13}{2024-01-23}{}
Continuation of sheet 8, exercise 4.
% Whiteboard https://wbo.ophir.dev/boards/Dphc7eylJJcIA0WbQsn7jzec1domqyx51gXb5qe6rzw-#263,0,0.5
\begin{definition}
Let $X$ be a compact metric space.
For $K \subseteq X$ compact and $U \overset{\text{open}}{\subseteq} X$
let
\[
S_{K,U} \coloneqq \{f \in \cC(X,X): f(K) \subseteq U\}.
\]
The \vocab{compact open topology} on $\cC(X,X)$
is the topology that has $S_{K,U}$ as a subbase.
\end{definition}
\begin{fact}
If $X$ is compact,
then the compact open topology
is the topology induced by the uniform metric $d_\infty$.
\end{fact}
\begin{proof}
Take some $S_{K,U}$. We need to show that this can be written
as a union of open $d_{\infty}$-balls.
Let $f_0 \in S_{K,U}$.
Consider the continuous function $d(-, U^c)$.
Since $f_0(K)$ is compact,
there exists $\epsilon \coloneqq \min d(f_0(K), U^c)$
and $B_{\epsilon}(f_0) \subseteq S_{K,U}$.
On the other hand, consider $B_{\epsilon}(f_0)$ for some $\epsilon > 0$
and $f_0 \in \cC(X,X)$.
As $f_0$ is uniformly continuous,
there exists $\delta > 0$ such that $d(x,x') < \delta \implies d(f_0(x), f_0(x')) < \frac{\epsilon}{3}$.
Cover $X$ with finitely many $\delta$-balls $B_\delta(a_1), \ldots, B_{\delta}(a_k)$.
Then
\[f_0(\overline{B_{\delta}(a_i)}) \subseteq \overline{f_0(B_{\delta}(a_i)} \subseteq \overline{B_{\frac{\epsilon}{3}}(f_0(a_i))} \subseteq B_{\frac{\epsilon}{2}}(f_0(a_i)).\]
For $i \le k$, let $S_i \coloneqq S_{\overline{B_{\delta}(a_i)}, B_{\frac{\epsilon}{2}}(f_0(a_i))}$.
Take $\bigcap_{i \le k} S_i$. This is open
in the compact open topology and
$B_{\epsilon}(f_0) \subseteq \bigcap_{i \le k} S_i$.
\end{proof}
\begin{claim}
$f \in \cC(X,X)$ is surjective
iff for all basic open $\emptyset\neq U \subseteq X$
there exists a basic open $\emptyset \neq V \subseteq X$
with $f(\overline{V}) \subseteq U$.
Note that we can write this as a $G_\delta$-condition.
\end{claim}
\begin{subproof}
Take $B_\epsilon(f(x_0))\subseteq U$.
Then there exists $\delta > 0$
such that $f(B_{\delta}(x_0)) \subseteq B_{\frac{\epsilon}{2}}(f(x_0))$
hence $f(\overline{B_{\delta}(x_0)}) \subseteq B_\epsilon(f(x_0))$.
For the other direction take $y \in X$.
We want to find a preimage.
For every $B_{\frac{1}{n}}(y)$,
there exists a basic open set $V_n$ with $f(\overline{V}) \subseteq B_{\frac{1}{n}}(y)$.
Take $x_n \in V_n$.
Since $X$ is compact, it is sequentially compact,
so there exists a converging subsequence.
Wlog.~$x_n \to x$,
so $f(x_n) \to f(x) = y$.
\end{subproof}
\begin{claim}
$f \in \cC(X,X)$ is injective iff
for all basic open $U$,$V$
with $\overline{U} \cap \overline{V} = \emptyset$
we have $f(\overline{U}) \cap f(\overline{V}) = \emptyset$.
This is a $G_\delta$-condition,
since we can write it as
\[
\bigcap_{U,V} S_{\overline{U}, f(\overline{V})^c}.
\]
\end{claim}
\begin{subproof}
$\implies$ is trivial.
$\impliedby$ follows since for all pairs $x,y \in X$,
we can find $x \in U$, $y \in V$ such that $\overline{U} \cap \overline{V} = \emptyset$.
\end{subproof}
Hence $\Homeo(X,X)$ is $G_\delta$.
In particular it is a Polish space.
Let $D$ be the set of $\Z$-flows with dense orbit.
\begin{claim}
$f \in D$ $\iff$
for all basic open $U,V \subseteq X$,
there exists $n \in \Z$
such that $f^n(U) \cap V \neq \emptyset$.
\end{claim}
\begin{subproof}
Suppose that the orbit of $x_0 \in X$ is dense.
Then there exist $k,l \in \Z$
such that $f^k(x_0)\in U$ and $f^l(x_0) \in V$,
so $f^{l-k} U \cap V \neq \emptyset$.
For basic open sets $V$
let
\[
A_V \coloneqq \{ x \in X: \exists n.~ f^n(x) \in V\}.
\]
By assumption, all the $A_V$ are dense.
Hence $\bigcap_{V}A_V$ is dense by the \yaref{thm:bct}.
$A_V = \bigcup_{n \in \Z} f^n(V)$ is open.
\end{subproof}
\begin{claim}
The condition can be written as a $G_\delta$ set.
\end{claim}
\begin{subproof}
For $f_0(U) \cap V \neq \emptyset$
take $u \in U$ such that $f_0(u) \in V$.
Then there exists $\epsilon > 0$ such that $B_{\epsilon}(f_0(u)) \subseteq U$,
hence $B_{\epsilon}(f_0)$ is an open neighbourhood contained
in $\{f : f(U) \cap V \neq \emptyset \} $.
For $n = 2$ note that
$d(f^2(u), f^2_0(u) \le d(f(f(u)), f_0(f(u))) + d(f_0(f(u)), f_0(f_0(u)))$.
The first part can be bounded by $d(f,f_0)$.
For the second part,
note that there exists $\delta$ such that
\[d(a,b) < \delta \implies d(f_0(a), f_0(b)) < \frac{\epsilon}{2}.\]
Let $\eta \coloneqq \min \{\delta, \frac{\epsilon}{2}\}$
and consider $d_\infty(f,f_0) < \epsilon$.
For other $n$ it is some more work, which is left as an exercise.
\end{subproof}

4
jrpie-gist.sty
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@ -1,6 +1,10 @@
\NeedsTeXFormat{LaTeX2e} \NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{jrpie-gist}[2023/01/22 - gist version for lecture notes] \ProvidesPackage{jrpie-gist}[2023/01/22 - gist version for lecture notes]
% TODO gist info
% TODO link to long version (provide link to main document)
% TODO \phantomsection to cross link
\newcommand{\gist}[2]{% \newcommand{\gist}[2]{%
\ifcsname EnableGist\endcsname% \ifcsname EnableGist\endcsname%
#2% #2%

1
logic3.tex
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@ -66,6 +66,7 @@
\input{inputs/tutorial_07} \input{inputs/tutorial_07}
\input{inputs/tutorial_08} \input{inputs/tutorial_08}
\input{inputs/tutorial_09} \input{inputs/tutorial_09}
\input{inputs/tutorial_13} % sic!
\input{inputs/tutorial_10} \input{inputs/tutorial_10}
\input{inputs/tutorial_11} \input{inputs/tutorial_11}
\input{inputs/tutorial_12b} \input{inputs/tutorial_12b}