diff git a/inputs/lecture_01.tex b/inputs/lecture_01.tex
index 7aebc22..df7eb8c 100644
 a/inputs/lecture_01.tex
+++ b/inputs/lecture_01.tex
@@ 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.
diff git a/inputs/lecture_02.tex b/inputs/lecture_02.tex
index 6faa1f3..01655a2 100644
 a/inputs/lecture_02.tex
+++ b/inputs/lecture_02.tex
@@ 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}

diff git a/inputs/lecture_03.tex b/inputs/lecture_03.tex
index e376cbe..d173a04 100644
 a/inputs/lecture_03.tex
+++ b/inputs/lecture_03.tex
@@ 2,7 +2,7 @@
\subsection{Trees}

+\gist{%
\begin{notation}
Let $A \neq \emptyset$, $n \in \N$.
Then
@@ 34,7 +34,7 @@
$s\defon{m} \coloneqq (s_0,\ldots,s_{m1})$.
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 $s = (s_0,\ldots, s_{n1})$
+ The \vocab{concatenation}
+ of $s = (s_0,\ldots, s_{n1})$
and $t = (t_0,\ldots, t_{m1})$
is the sequence
$s\concat t \coloneqq (s_0,\ldots,s_{n1}, t_0,\ldots, t_{n1})$
@@ 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 nonempty open set
 with small enough diameter.
+ \gist{%
+ For $U_{\emptyset}$ take any nonempty 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 $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i1})$
are $F_\sigma$, disjoint
and $F_i^0 = \bigcup_{j} D_j$.
+}{Induction.}
diff git a/inputs/lecture_04.tex b/inputs/lecture_04.tex
index ff9fbb8..4c16ba2 100644
 a/inputs/lecture_04.tex
+++ b/inputs/lecture_04.tex
@@ 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}
+}
diff git a/inputs/lecture_05.tex b/inputs/lecture_05.tex
index e2c10a1..748b6a1 100644
 a/inputs/lecture_05.tex
+++ b/inputs/lecture_05.tex
@@ 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}
+}{}
diff git a/inputs/lecture_06.tex b/inputs/lecture_06.tex
index 1f74848..f3bb7e0 100644
 a/inputs/lecture_06.tex
+++ b/inputs/lecture_06.tex
@@ 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 $\lnot$ (elementwise 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,
$\Sigma^0_2 = F_\sigma$,
diff git a/inputs/lecture_07.tex b/inputs/lecture_07.tex
index 687e009..22a2026 100644
 a/inputs/lecture_07.tex
+++ b/inputs/lecture_07.tex
@@ 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 nonstandard 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.
diff git a/inputs/lecture_08.tex b/inputs/lecture_08.tex
index 4288b7f..62298a0 100644
 a/inputs/lecture_08.tex
+++ b/inputs/lecture_08.tex
@@ 1,61 +1,8 @@
\lecture{08}{20231110}{}

\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}{20231110}{}\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}
diff git a/inputs/tutorial_01.tex b/inputs/tutorial_01.tex
index 2ef9ca6..3eb6d60 100644
 a/inputs/tutorial_01.tex
+++ b/inputs/tutorial_01.tex
@@ 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}