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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}
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
closely. Additions (e.g.~from exercise sheets)
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
\[
\tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}.
\]
\]
Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
\end{example}
\begin{example}[Sorgenfrey line]
@ -76,7 +76,7 @@ However the converse of this does not hold.
\end{itemize}
\end{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
by open sets.
\end{fact}
@ -114,7 +114,7 @@ However the converse of this does not hold.
\end{absolutelynopagebreak}
\subsection{Some facts about polish spaces}
\gist{%
\begin{fact}
Let $(X, \tau)$ be a topological space.
Let $d$ be a metric on $X$.
@ -130,9 +130,10 @@ To show that $\tau_d = \tau_{d'}$
for two metrics $d, d'$,
suffices to show that open balls in one metric are unions of open balls in the other.
\end{fact}
}{}
\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}
\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 $\tau$.
\end{proposition}
\gist{%
\begin{proof}
To check the triangle inequality:
\begin{IEEEeqnarray*}{rCl}
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).
\end{IEEEeqnarray*}
For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$
and for $\epsilon > 1$, $B'_\epsilon(x) = X$.
Since $d$ is complete, we have that $d'$ is complete.
\end{proof}
}{}
\begin{proposition}
Let $A$ be a Polish space.
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).
Let $d \le 1$ be a complete metric on $A$.
Define $D$ on $A^\omega$ by
Define $D$ on $A^\omega$ by
\[
D\left( (x_n), (y_n) \right) \coloneqq
D\left( (x_n), (y_n) \right) \coloneqq
\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
\]
\]
Clearly $D \le 1$.
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.}.
Then $X$ topologically embeds into the
\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.
\end{proposition}
\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}
Closed subspaces of Polish spaces are Polish.
\end{proposition}
\gist{}{
\gist{%
\begin{proof}
Let $X$ be Polish and $V \subseteq X$ closed.
Let $d$ be a complete metric on $X$.
Then $d\defon{V}$ is complete.
Subspaces of second countable spaces
are second countable.
\end{proof}
}
\end{proof}%
}{}
\begin{definition}
Let $X$ be a topological space.

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

86
inputs/lecture_03.tex
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@ -2,7 +2,7 @@
\subsection{Trees}
\gist{%
\begin{notation}
Let $A \neq \emptyset$, $n \in \N$.
Then
@ -34,7 +34,7 @@
$s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$.
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 $s$)
if there exists an $n$ such that $s = t\defon{|s|}$.
We write this as $s \subseteq t$.
@ -44,8 +44,8 @@
Otherwise the are \vocab{incompatible},
we denote that as $s \perp t$.
The \vocab{concatenation}
of $s = (s_0,\ldots, s_{n-1})$
The \vocab{concatenation}
of $s = (s_0,\ldots, s_{n-1})$
and $t = (t_0,\ldots, t_{m-1})$
is the sequence
$s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$
@ -59,10 +59,11 @@
define extension, initial segments
and concatenation of a finite sequence with an infinite one.
\end{notation}
}{}
\begin{definition}
A \vocab{tree}
on a set $A$ is a subset $T \subseteq A^{<\N}$
A \vocab{tree}
on a set $A$ is a subset $T \subseteq A^{<\N}$
closed under initial segments,
i.e.~if $t \in T, s \subseteq t \implies s \in T$.
Elements of trees are called \vocab{nodes}.
@ -70,27 +71,27 @@
A \vocab{leave} is an element of $T$,
that has no extension in $t$.
An \vocab{infinite branch} of a tree $T$
is $x \in A^{\N}$
An \vocab{infinite branch} of a tree $T$
is $x \in A^{\N}$
such that $\forall n.~x\defon{n} \in T$.
The \vocab{body} of $T$ is the set of all
infinite branches:
\[
[T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.
\]
\]
We say that $T$ is \vocab{pruned},
iff
\[
\forall t\in T.\exists s \supsetneq t.~s \in T.
\]
\]
\end{definition}
\begin{definition}
A \vocab{Cantor scheme}
on a set $X$ is a family
$(A_s)_{s \in 2^{< \N}}$
on a set $X$ is a family
$(A_s)_{s \in 2^{< \N}}$
of subsets of $X$ such that
\begin{enumerate}[i)]
\item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
@ -99,7 +100,7 @@
\end{definition}
\begin{definition}
A topological space
A topological space
is \vocab{perfect}
if it has no isolated points,
i.e.~for any $U \neq \emptyset$ open,
@ -108,15 +109,15 @@
\begin{theorem}
\label{thm:cantortopolish}
Let $X \neq \emptyset$
Let $X \neq \emptyset$
be a perfect Polish space.
Then there is an embedding
of the Cantor space $2^{\N}$
of the Cantor space $2^{\N}$
into $X$.
\end{theorem}
\begin{proof}
We will define a Cantor scheme
$(U_s)_{s \in 2^{<\N}}$
$(U_s)_{s \in 2^{<\N}}$
such that $\forall s \in 2^{< \N}$.
\begin{enumerate}[(i)]
\item $U_s \neq \emptyset$ and open,
@ -127,39 +128,46 @@
We define $U_s$ inductively on the length of $s$.
For $U_{\emptyset}$ take any non-empty open set
with small enough diameter.
\gist{%
For $U_{\emptyset}$ take any non-empty open set
with small enough diameter.
Given $U_s$, pick $x \neq y \in U_s$
and let $U_{s \concat 0} \ni x$,
$U_{s \concat 1} \ni y$
be disjoint, open,
of diameter $\le \frac{1}{2^{|s| +1}}$
and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
Given $U_s$, pick $x \neq y \in U_s$
and let $U_{s \concat 0} \ni x$,
$U_{s \concat 1} \ni y$
be disjoint, open,
of diameter $\le \frac{1}{2^{|s| +1}}$
and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
}{}
\gist{%
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
\[
\{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}.
\]
\]
(This is nonempty as $X$ is a completely metrizable space.)
It is clear that $f$ is injective and continuous.
% TODO: more details
$2^{\N}$ 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}
\begin{corollary}
\label{cor:perfectpolishcard}
Every nonempty perfect Polish
space $X$ has cardinality $\fc = 2^{\aleph_0}$
% TODO: eulerscript C ?
space $X$ has cardinality $\fc = 2^{\aleph_0}$
\end{corollary}
\begin{proof}
Since the cantor space embeds into $X$,
we get the lower bound.
Since $X$ is second countable and Hausdorff,
we get the upper bound.
\gist{%
Since the cantor space embeds into $X$,
we get the lower 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}
\begin{theorem}
@ -173,13 +181,13 @@
\begin{definition}
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 $
such that
\begin{enumerate}[(i)]
\item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$
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 $i \in \N, s \in \N^{<\N}$.
\end{enumerate}
\end{definition}
@ -191,11 +199,11 @@
\[
D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN
% TODO correct N for the Baire space?
\]
\]
and a continuous bijection from
$D$ 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 $f$.
\end{theorem}
\begin{definition}
@ -203,12 +211,14 @@
countable union of closed sets,
i.e.~the complement of a $G_\delta$ set.
\end{definition}
\gist{%
\begin{observe}
\begin{itemize}
\item Any open set is $F {\sigma}$.
\item In metric spaces the intersection of an open and closed set is $F_\sigma$.
\end{itemize}
\end{observe}
}{}
\begin{refproof}{thm:bairetopolish}
Let $d$ be a complete metric on $X$.
W.l.o.g.~$\diam(X) \le 1$.
@ -220,7 +230,7 @@
\item $F_\emptyset = X$,
\item $F_s$ is $F_\sigma$ for all $s$.
\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
$\overline{F_{s \concat i}} \subseteq F_s$
@ -228,6 +238,7 @@
\item $\diam(F_s) \le 2^{-|s|}$.
\end{enumerate}
\gist{%
Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
We need to construct a partition $(F_i)_{i \in \N}$
of $F$ with $\overline{F_i} \subseteq F$
@ -252,6 +263,7 @@
The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
are $F_\sigma$, disjoint
and $F_i^0 = \bigcup_{j} D_j$.
}{Induction.}

74
inputs/lecture_04.tex
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@ -5,6 +5,7 @@
\end{remark}
\begin{refproof}{thm:bairetopolish}
\gist{%
Take
\[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
@ -12,16 +13,19 @@
we have
\[
\bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
\]
\]
}{}
$f\colon D \to X$ is determined by
\[
\{f(x)\} = \bigcap_{n} F_{x\defon{n}}
\]
\]
$f$ is injective and continuous.
The proof of this is exactly the same as in
\yaref{thm:cantortopolish}.
\gist{%
$f$ is injective and continuous.
The proof of this is exactly the same as in
\yaref{thm:cantortopolish}.
}{}
\begin{claim}
\label{thm:bairetopolish:c1}
@ -47,7 +51,7 @@
there exists $y = \lim_n f(x_n)$.
Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
we get that $y \in \overline{F_{x\defon{N}}}$.
Note that for $N' > N$ by the same argument
we get $y \in \overline{F_{x\defon{N'}}}$.
Hence
@ -55,20 +59,20 @@
i.e.~$y \in D$ and $y = f(x)$.
\end{refproof}
We extend $f$ to $g\colon\cN \to X$
We extend $f$ to $g\colon\cN \to X$
in the following way:
Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
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})
\[
D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
\]
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).
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
such that
\begin{itemize}
@ -76,21 +80,27 @@
\item $|s| = \phi(|s|)$,
\item if $s \in S$, then $\phi(s) = s$.
\end{itemize}
Let $\phi(\emptyset) = \emptyset$.
Suppose that $\phi(t)$ is defined.
If $t\concat a \in S$, then set
$\phi(t\concat a) \coloneqq t\concat a$.
Otherwise take some $b$ such that
$t\concat b \in S$ and define
$\phi(t\concat a) \coloneqq \phi(t)\concat b$.
\gist{%
Let $\phi(\emptyset) = \emptyset$.
Suppose that $\phi(t)$ is defined.
If $t\concat a \in S$, then set
$\phi(t\concat a) \coloneqq t\concat a$.
Otherwise take some $b$ such that
$t\concat b \in S$ and define
$\phi(t\concat a) \coloneqq \phi(t)\concat b$.%
}{}%
This is possible since $S$ is pruned.
Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
\gist{%
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
$d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz
It is immediate that $r$ is a retraction.
\gist{%
It is immediate that $r$ is a retraction.
}{}
\end{refproof}
\section{Meager and Comeager Sets}
@ -110,42 +120,48 @@
then $A \cap U$ is not dense in $U$).
\end{itemize}
A set $B \subseteq X$ is \vocab{meager}
A set $B \subseteq X$ is \vocab{meager}
(or \vocab{first category}),
iff it is a countable union of nwd sets.
The complement of a meager set is called
\vocab{comeager}.
\end{definition}
\gist{%
\begin{example}
$\Q \subseteq \R$ is meager.
\end{example}
}{}
\begin{notation}
Let $A, B \subseteq X$.
We write $A =^\ast B$
We write $A =^\ast B$
iff the \vocab{symmetric difference},
$A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$,
is meager.
\end{notation}
\gist{%
\begin{remark}
$=^\ast$ is an equivalence relation.
\end{remark}
}{}
\begin{definition}
A set $A \subseteq X$
has the \vocab{Baire property} (\vocab{BP})
if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
\end{definition}
\gist{%
Note that open sets and meager sets have the Baire property.
}{}
\gist{%
\begin{example}
\begin{itemize}
\item $\Q \subseteq \R$ is $F_\sigma$.
\item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
\item $\Q \subseteq \R$ is not $G_{\delta}$.
(It is dense and meager,
\item $\Q \subseteq \R$ is not $G_{\delta}$:
It is dense and meager,
hence it can not be $G_\delta$
by the Baire category theorem).
by the \yaref{thm:bct}.
\end{itemize}
\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 Any meager set $B$ is contained in a meager $F_{\sigma}$-set.
\end{itemize}
\end{fact}
\begin{proof} % remove?
\gist{%
\begin{proof}
\begin{itemize}
\item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.
\item Trivial.
@ -17,7 +16,9 @@
Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.
\end{itemize}
\end{proof}
}{}
\gist{%
\begin{definition}
A \vocab{$\sigma$-algebra} on a set $X$
is a collection of subsets of $X$
@ -32,14 +33,15 @@
Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$
we have that $\sigma$-algebras are closed under countable intersections.
\end{fact}
}{}
\begin{theorem}
\label{thm:bairesigma}
Let $X$ be a topological space.
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.
\end{theorem}
\begin{refproof}{thm:bairesigma}
@ -93,10 +95,11 @@
% Nwd set of positive measure.
% TODO
% remove open intervals such that their length does not add to 0
%
%
%\end{example}
\begin{theorem}[Baire Category theorem]
\yalabel{Baire Category Theorem}{Baire Category Theorem}{thm:bct}
Let $X$ be a completely metrizable space.
Then every comeager set of $X$ is dense in $X$.
\end{theorem}
@ -111,8 +114,8 @@
\item The intersection of countable many
open dense sets is dense.
\end{enumerate}
In this case $X$ is called a \vocab{Baire space}.
\footnote{see \yaref{s5e1}}
In this case $X$ is called a \vocab{Baire space}.%
\footnote{cf.~\yaref{s5e1}}
\end{theoremdef}
\begin{proof}
\todo{Proof (short)}
@ -124,12 +127,11 @@
We have that
\[
\emptyset = \bigcap_{n} (X \setminus \overline{A_n})
\]
\]
is dense by (iii).
This proof can be adapted to other open sets $X$.
\end{proof}
\begin{notation}
Let $X ,Y$ be topological spaces,
$A \subseteq X \times Y$
@ -138,12 +140,12 @@
Let
\[
A_x \coloneqq \{y \in Y : (x,y) \in A\}
\]
A_x \coloneqq \{y \in Y : (x,y) \in A\}
\]
and
\[
A^y \coloneqq \{x \in X : (x,y) \in A\} .
\]
\]
\end{notation}
The following similar to Fubini,
@ -273,9 +275,11 @@ but for meager sets:
% \end{refproof}
% TODO fix claim numbers
\gist{%
\begin{remark}
Suppose that $A$ has the BP.
Then there is an open $U$ such that
$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
Then $A = U \symdif M$.
\end{remark}
}{}

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

78
inputs/lecture_07.tex
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@ -16,17 +16,17 @@
We have that
\[
\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
\]
\]
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
\[
\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).
\]
\]
Hence
\[
|\cB(X)| \le \omega_1 \cdot \fc = \fc.
\]
\]
\end{proof}
\begin{proposition}[Closure properties]
@ -37,54 +37,58 @@
\item \begin{itemize}
\item $\Sigma^0_\xi(X)$ is closed under countable unions.
\item $\Pi^0_\xi(X)$ is closed under countable intersections.
\item $\Delta^0_\xi(X)$ is closed under complements,
countable unions and
countable intersections.
\item $\Delta^0_\xi(X)$ is closed under complements.
\end{itemize}
\item \begin{itemize}
\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
\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{enumerate}
\end{proposition}
\gist{%
\begin{proof}
\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
as a complement of the countable union of complements:
\[
\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
\]
\]
\item If suffices to check this for $\Sigma^0_{\xi}(X)$.
Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
Then
\[
A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)
\]
\]
and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
\end{enumerate}
\end{proof}
}{}
\begin{example}
Consider the cantor space $2^{\omega}$.
We have that $\Delta^0_1(2^{\omega})$
is not closed under countable unions
(countable unions yield all open sets, but there are open
sets that are not clopen).
is not closed under countable unions%
\gist{ (countable unions yield all open sets, but there are open
sets that are not clopen)}{}.
\end{example}
\subsection{Turning Borels Sets into Clopens}
\begin{theorem}%
\gist{%
\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
unfortunately seems to be non-standard vocabulary.
Our tutor repeatedly advised against using it in the final exam.
Contrary to popular belief
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.
}
}%
}{}%
\label{thm:clopenize}
Let $(X, \cT)$ be a Polish space.
For any Borel set $A \subseteq X$,
@ -163,7 +167,7 @@
such that $\cT_n \supseteq \cT$
and $\cB(\cT_n) = \cB(\cT)$.
Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
is still Polish
is Polish
and $\cB(\cT_\infty) = \cB(T)$.
\end{lemma}
\begin{proof}
@ -183,7 +187,51 @@
definition of $\cF$ belong to
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}
We need to show that $A$ is closed under countable unions.

63
inputs/lecture_08.tex
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@ -1,61 +1,8 @@
\lecture{08}{2023-11-10}{}
\todo{put this lemma in the right place}
\begin{lemma}[Lemma 2]
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}
\lecture{08}{2023-11-10}{}\footnote{%
In the beginning of the lecture, we finished
the proof of \yaref{thm:clopenize:l2}.
This has been moved to the notes on lecture 7.%
}
\subsection{Parametrizations}
%\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}
\label{thm:universals11}
Let $X,Y$ be uncountable Polish spaces.
There exists a $Y$-universal $\Sigma^1_1(X)$ set.
\end{theorem}

20
inputs/lecture_11.tex
View File

@ -153,13 +153,14 @@ We will not proof this in this lecture.
\subsection{Ill-Founded Trees}
\gist{%
Recall that a \vocab{tree} on $\N$ is a subset of
$\N^{<\N}$
closed under taking initial segments.
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}
\One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\
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.
\end{cases}
\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}
\[
\Tr \subseteq {2^{\N}}^{<\N}
\]
is closed (where we take the topology of the Cantor space).
$\Tr \subseteq {2^{\N}}^{<\N}$ is closed
(where we take the topology of the Cantor space).
\end{observe}
\gist{%
Indeed, for any $ s \in \N^{<\N}$
we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
@ -185,6 +185,4 @@ In particular for $s$ fixed,
we have that
\[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
is clopen in ${2^{\N}}^{<\N}$.
}{}

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

43
inputs/lecture_13.tex
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@ -1,11 +1,12 @@
\lecture{13}{2023-11-08}{}
\gist{%
% Recap
$\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
$\LO \subseteq 2^{\N \times \N}$ is closed
and $\WO = \{x \in \LO: x \text{ is a wellordering}\} $
is coanalytic in $\LO$.
% End Recap
}{}
Another way to code linear orders:
@ -32,6 +33,7 @@ with $(f^{-1}(\{1\}), <)$.
and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
\end{proof}
\begin{definition}[\vocab{Kleene-Brouwer ordering}]
Let $(A,<)$ be a linear order and $A$ countable.
We define the linear order $<_{KB}$ on $A^{<\N}$
@ -55,25 +57,30 @@ with $(f^{-1}(\{1\}), <)$.
$(T, <_{KB}\defon{T})$ is well ordered.
\end{proposition}
\begin{proof}
If $T$ is ill-founded and $x \in [T]$,
then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
Thus $(T, <_{KB}\defon{T})$ is not well ordered.
\gist{%
If $T$ is ill-founded and $x \in [T]$,
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
on $T$.
Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
be an infinite descending chain.
We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
stabilizes for $n > n_0$.
Let $a_0 \coloneqq s_{n_0}(0)$.
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.
Thus there is $n_1 \ge n_0$ such that $s_n(1)$
is constant for all $n \ge n_1$.
Let $a_1 \coloneqq s_{n_1}(1)$
and so on.
Then $(a_0,a_1,a_2, \ldots) \in [T]$.
Conversely, let $<\defon{KB}$ be not a well-ordering
on $T$.
Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
be an infinite descending chain.
We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
stabilizes for $n > n_0$.
Let $a_0 \coloneqq s_{n_0}(0)$.
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.
Thus there is $n_1 \ge n_0$ such that $s_n(1)$
is constant for all $n \ge n_1$.
Let $a_1 \coloneqq s_{n_1}(1)$
and so on.
Then $(a_0,a_1,a_2, \ldots) \in [T]$.
}{easy}
\end{proof}
% TODO ANKI-MARKER
\begin{theorem}[Lusin-Sierpinski]
The set $\LO \setminus \WO$
(resp.~$2^{\Q} \setminus \WO$)

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

11
inputs/lecture_16.tex
View File

@ -175,9 +175,10 @@ By Zorn's lemma, this will follow from
\end{IEEEeqnarray*}
The topology induced by the metric
is given by basic open subsets\footnote{see Exercise Sheet 9% TODO REF
}
$[U_0; U_1,\ldots, U_n]$, $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
is given by basic open subsets\footnote{cf.~\yaref{s9e2}}
of the form
$[U_0; U_1,\ldots, U_n]$,
for $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
where
\[
[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.
For a fixed compact metric space $X$,
we can view the flows $(X,\Z)$ as a subset of $\cC(X,X)$.
Note that $\cC(X,X)$ is Polish.
Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.
Note that $\cC(X,X)$ is Polish.\footnote{cf.~\yaref{s1e4}}
Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.\footnote{Exercise} % TODO
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.

13
inputs/tutorial_01.tex
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@ -27,8 +27,10 @@
\begin{definition}
A topological space is \vocab{Lindelöf}
if every open cover has a countable subcover.
iff every open cover has a countable subcover.
\end{definition}
\begin{fact}
Let $X$ be a metric space.
If $X$ is Lindelöf,
@ -64,5 +66,12 @@
and Lindelöf coincide.
In arbitrary topological spaces,
Lindelöf is the strongest of these notions.
Lindelöf is the weakest of these notions.
\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{IEEEeqnarray*}
Clearly this has the desired properties.
\item We have
\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.
\item Let $X$ be a compact Polish space.
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.
@ -160,14 +151,9 @@ amounts to a finite number of conditions on the preimage.
\end{IEEEeqnarray*}
by the general 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.
Then $f$ is a homeomorphism.
\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}

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}
\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]{%
\ifcsname EnableGist\endcsname%
#2%

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