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gist for lectures 1-4
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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}
}

14
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}
@ -274,9 +276,11 @@ 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}

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}