gist for lectures 1-4

inputs/lecture_01.tex

@@ -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}
@@ -142,6 +143,7 @@ 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}
@@ -154,6 +156,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
 
     Since $d$ is complete, we have that $d'$ is complete.
 \end{proof}
+}{}
 \begin{proposition}
     Let $A$ be a Polish space.
     Then $A^{\omega}$ Polish.
@@ -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.

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{:
+        $f_U$ is an embedding of $U$ into $X \times \R$%
-        \begin{itemize}
+        \gist{:
-            \item It is injective because of the first coordinate.
+            \begin{itemize}
-            \item It is continuous since $d(x, U^c)$ is continuous
+                \item It is injective because of the first coordinate.
-                and only takes strictly positive values. % TODO
+                \item It is continuous since $d(x, U^c)$ is continuous
-            \item The inverse is continuous because projections
+                    and only takes strictly positive values. % TODO
-                are continuous.
+                \item The inverse is continuous because projections
-        \end{itemize}
+                    are continuous.
-    }{.}
+            \end{itemize}
+        }{.}
 
-        So we have shown that $U$ and
+        \gist{%
-        the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
+            So we have shown that $U$ and
-        are homeomorphic.
+            the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
-        The graph is closed \gist{in $U \times  \R$,
+            are homeomorphic.
-        because $\tilde{f_U}$ is continuous.
+            The graph is closed \gist{in $U \times  \R$,
-        It is closed}{} in $X \times \R$ \gist{because
+            because $\tilde{f_U}$ is continuous.
-        $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
+            It is closed}{} in $X \times \R$ \gist{because
-        \todo{Make this precise}
+            $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
+            \todo{Make this precise}
+
+            Therefore we identified $U$ with a closed subspace of
+            the Polish space $(X \times  \R, d_1)$.
+        }{%
+            So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$
+            and the RHS is a close subspace of the Polish space
+            $(X \times \R, d_1)$.
+        }
 
-        Therefore we identified $U$ with a closed subspace of
-        the Polish space $(X \times  \R, d_1)$.
     \end{refproof}
 
     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.
+    \gist{
-    Since $X \times \R^{\N}$
+        As for an open $U$, $f_Y$ is an embedding.
-    is completely metrizable,
+        Since $X \times \R^{\N}$
-    so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
+        is completely metrizable,
+        so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
+    }{}
 
     \begin{claim}
         \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{
+        \gist{%
-        We want to show that $Y = \bigcap_{n \in \N} V_n$.
+            We want to show that $Y = \bigcap_{n \in \N} V_n$.
-        For $x \in Y$, $n \in \N$ we have $x \in V_n$,
+            For $x \in Y$, $n \in \N$ we have $x \in V_n$,
-        as we can choose two neighbourhoods
+            as we can choose two neighbourhoods
-        $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
+            $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
-        such that $\diam_{d_Y}(U) < \frac{1}{n}$
+            such that $\diam_{d_Y}(U) < \frac{1}{n}$
-        and $U_2 \cap Y = U_1$.
+            and $U_2 \cap Y = U_1$.
-        Additionally choose $x \in U_3$ open in $X$
+            Additionally choose $x \in U_3$ open in $X$
-        with $\diam_{d}(U_3) < \frac{1}{n}$.
+            with $\diam_{d}(U_3) < \frac{1}{n}$.
-        Then consider $U_2 \cap U_3 \subseteq V_n$.
+            Then consider $U_2 \cap U_3 \subseteq V_n$.
-        Hence $Y \subseteq  \bigcap_{n \in \N}  V_n$.
+            Hence $Y \subseteq  \bigcap_{n \in \N}  V_n$.
 
-        Now let $x \in \bigcap_{n \in \N} V_n$.
+            Now let $x \in \bigcap_{n \in \N} V_n$.
-        For each $n$ pick $x \in U_n \subseteq X$ open
+            For each $n$ pick $x \in U_n \subseteq X$ open
-        satisfying (i), (ii), (iii).
+            satisfying (i), (ii), (iii).
-        From (i) and (ii) it follows that $x \in \overline{Y}$,
+            From (i) and (ii) it follows that $x \in \overline{Y}$,
-        since we can consider a sequence of points $y_n \in U_n \cap Y$
+            since we can consider a sequence of points $y_n \in U_n \cap Y$
-        and get $y_n \xrightarrow{d} x$.
+            and get $y_n \xrightarrow{d} x$.
-        For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
+            For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
-        is an open set containing $x$,
+            is an open set containing $x$,
-        hence $U_n' \cap Y \neq \emptyset$.
+            hence $U_n' \cap Y \neq \emptyset$.
-        Thus we may assume that the $U_i$ form a decreasing sequence.
+            Thus we may assume that the $U_i$ form a decreasing sequence.
-        We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
+            We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
-        If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
+            If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
-        since $\diam(U_n \cap Y) \xrightarrow{d_Y}  0$
+            since $\diam(U_n \cap Y) \xrightarrow{d_Y}  0$
-        and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y}  0$.
+            and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y}  0$.
-        The sequence $y_n$ converges to the unique point in
+            The sequence $y_n$ converges to the unique point in
-        $\bigcap_{n} \overline{U_n \cap Y}$.
+            $\bigcap_{n} \overline{U_n \cap Y}$.
-        Since the topologies agree, this point is $x$.
+            Since the topologies agree, this point is $x$.
-    }{Then $Y = \bigcap_n U_n$.}
+        }{Then $Y = \bigcap_n U_n$.}
     \end{refproof}
 \end{refproof}
-

inputs/lecture_03.tex

@@ -2,7 +2,7 @@
 
 \subsection{Trees}
 
-
+\gist{%
 \begin{notation}
     Let $A \neq  \emptyset$, $n \in \N$.
     Then
@@ -59,6 +59,7 @@
     define extension, initial segments
     and concatenation of a finite sequence with an infinite one.
 \end{notation}
+}{}
 
 \begin{definition}
     A \vocab{tree}
@@ -127,16 +128,19 @@
 
     We define $U_s$ inductively on the length of $s$.
 
-    For $U_{\emptyset}$ take any non-empty open set
+    \gist{%
-    with small enough diameter.
+        For $U_{\emptyset}$ take any non-empty open set
+        with small enough diameter.
 
-    Given $U_s$, pick $x \neq y \in U_s$
+        Given $U_s$, pick $x \neq y \in U_s$
-    and let $U_{s \concat 0} \ni x$,
+        and let $U_{s \concat 0} \ni x$,
-    $U_{s \concat 1} \ni y$ 
+        $U_{s \concat 1} \ni y$
-    be disjoint, open,
+        be disjoint, open,
-    of diameter $\le \frac{1}{2^{|s| +1}}$
+        of diameter $\le \frac{1}{2^{|s| +1}}$
-    and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq  U_s$.
+        and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq  U_s$.
+    }{}
 
+    \gist{%
     Let $x \in  2^{\N}$.
     Then let $f(x)$ be the unique point in $X$
     such that
@@ -147,19 +151,23 @@
     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 ?
 \end{corollary}
 \begin{proof}
-    Since the cantor space embeds into $X$,
+    \gist{%
-    we get the lower bound.
+        Since the cantor space embeds into $X$,
-    Since $X$ is second countable and Hausdorff,
+        we get the lower bound.
-    we get the upper bound.
+        Since $X$ is second countable and Hausdorff,
+        we get the upper bound.%
+    }{Lower bound:  $2^{\N} \hookrightarrow X$,
+        upper bound: \nth{2} countable and Hausdorff.
 \end{proof}
 
 \begin{theorem}
@@ -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.}
 
 
 

inputs/lecture_04.tex

@@ -5,6 +5,7 @@
 \end{remark}
 
 \begin{refproof}{thm:bairetopolish}
+    \gist{%
     Take
     \[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
 
@@ -13,15 +14,18 @@
     \[
     \bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
     \]
+    }{}
 
     $f\colon D \to  X$ is determined by
     \[
     \{f(x)\} = \bigcap_{n} F_{x\defon{n}}
     \]
 
-    $f$ is injective and continuous.
+    \gist{%
-    The proof of this is exactly the same as in 
+        $f$ is injective and continuous.
-    \yaref{thm:cantortopolish}.
+        The proof of this is exactly the same as in
+        \yaref{thm:cantortopolish}.
+    }{}
 
     \begin{claim}
         \label{thm:bairetopolish:c1}
@@ -60,7 +64,7 @@
 
     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})
     \[
     D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
     \]
@@ -76,21 +80,27 @@
         \item $|s| = \phi(|s|)$,
         \item if $s \in S$, then $\phi(s) = s$.
     \end{itemize}
-    Let $\phi(\emptyset) = \emptyset$.
+    \gist{%
-    Suppose that $\phi(t)$ is defined.
+        Let $\phi(\emptyset) = \emptyset$.
-    If $t\concat a \in S$, then set
+        Suppose that $\phi(t)$ is defined.
-     $\phi(t\concat a) \coloneqq t\concat a$.
+        If $t\concat a \in S$, then set
-    Otherwise take some $b$ such that
+         $\phi(t\concat a) \coloneqq t\concat a$.
-    $t\concat b \in S$ and define
+        Otherwise take some $b$ such that
-    $\phi(t\concat a) \coloneqq \phi(t)\concat b$.
+        $t\concat b \in S$ and define
+        $\phi(t\concat a) \coloneqq \phi(t)\concat b$.%
+    }{}%
     This is possible since $S$ is pruned.
 
-    Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
+    \gist{%
-    be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
+        Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
+        be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
+    }{}
 
     $r$ is continuous, since
     $d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % 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}
@@ -117,9 +127,11 @@
     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 $A =^\ast B$
@@ -127,25 +139,29 @@
     $A \symdif B \coloneqq  (A\setminus B) \cup (B \setminus A)$,
     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 $G_{\delta}$.
+        \item $\Q \subseteq \R$ is not $G_{\delta}$:
-            (It is dense and meager,
+            It is dense and meager,
             hence it can not be $G_\delta$
-            by the Baire category theorem).
+            by the \yaref{thm:bct}.
     \end{itemize}
-
 \end{example}
+}

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}
+}{}

inputs/lecture_06.tex

@@ -101,7 +101,6 @@ Then define
 \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 $\alpha_n < \alpha$}\}.

inputs/lecture_07.tex

@@ -37,17 +37,18 @@
         \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,
+                \item  $\Delta^0_\xi(X)$ is closed under complements.
-                    countable unions and
-                    countable intersections.
             \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.
@@ -67,24 +68,27 @@
 
     \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
+    is not closed under countable unions%
-    (countable unions yield all open sets, but there are open
+    \gist{ (countable unions yield all open sets, but there are open
-    sets that are not clopen).
+    sets that are not clopen)}{}.
 \end{example}
 
 \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.

inputs/lecture_08.tex

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

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}