some small changes

inputs/lecture_05.tex

@@ -164,8 +164,10 @@
     \]
 \end{notation}
 
+\gist{%
 The following similar to Fubini,
 but for meager sets:
+}{}
 
 \begin{theorem}[Kuratowski-Ulam]
     \yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam}
@@ -193,6 +195,7 @@ but for meager sets:
     \end{enumerate}
 \end{theorem}
 \begin{refproof}{thm:kuratowskiulam}
+\gist{
     (ii) and (iii) are equivalent by passing to the complement.
 
     \begin{claim}%[1a]
@@ -286,16 +289,11 @@ but for meager sets:
         $M_x$ is comeager
         as a countable intersection of comeager sets.
     \end{refproof}
+}{}
+
 
 % \phantom\qedhere
 % \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

@@ -1,8 +1,8 @@
 \lecture{06}{2023-11-03}{}
-
+\gist{%
 % \begin{refproof}{thm:kuratowskiulam}
     \begin{enumerate}[(i)]
-        \item Let $A$ be a set with the Baire Property.
+        \item Let $A$ be a set with the Baire property.
             Write $A = U \symdif M$
             for $U$ open and $M$ meager.
             Then for all $x$,
@@ -51,8 +51,8 @@
             Towards a contradiction suppose that  $A$ is not meager.
             Then $U$ is not meager.
             Since $X \times  Y$ is second countable,
-            we have that $A$ is a countable union of open rectangles.
-            At least one of them, say $G \times H \subseteq A$,
+            we have that $U$ is a countable union of open rectangles.
+            At least one of them, say $G \times H \subseteq U$,
             is not meager.
             By \yaref{thm:kuratowskiulam:c2},
             both $G$ and $H$ are not meager.
@@ -71,7 +71,59 @@
             ``$\implies$''
             This is \yaref{thm:kuratowskiulam:c1b}.
     \end{enumerate}
+}{%
+    \begin{itemize}
+        \item (ii) $\iff$ (iii): pass to complement.
+        \item $F \overset{\text{closed}}{\subseteq} X \times Y$ nwd.
+            $\implies \{x \in X : F_x \text{ nwd}\} $ comeager:
+            \begin{itemize}
+                \item $W = F^c$ is open and dense, show that $\{x : W_x \text{ dense}\}$
+                    is comeager.
+                \item $(V_n)$ enumeration of basis. Show that $U_n \coloneqq  \{x : V_n \cap W_x \neq \emptyset\}$
+                    is comeager for all $n$.
+                \item $U_n$ is open (projection of open) and dense ($W$ is dense, hence $W \cap ( U \times V_n) \neq \emptyset$ for $U$ open).
+            \end{itemize}
+        \item $F \subseteq X \times Y$ is nwd $\implies \{x \in X: F_x \text{ nwd}\}$ comeager.
+            (consider $\overline{F}$).
+        \item (ii) $\implies$:
+            $M \subseteq X \times Y$ meager $\implies \{x \in X: M_x \text{ meager}\}$ comeager
+            (write $M$ as ctbl. union of nwd.)
+        \item (i):  If $A$ has the Baire Property,
+            then $A = U \symdif M$, $A_x = U_x \symdif M_x$,
+             $U_x$ open and $\{x : M_x \text{ meager}\}$ comeager
+             $\implies$ (i).
+        \item $P \subseteq X$, $Q \subseteq Y$ BP,
+            then $P \times Q$ meager $\iff$ $P$ or $Q$ meager.
+            \begin{itemize}
+                \item $\impliedby$ easy
+                \item $\implies$ Suppose $P \times Q$ meager, $P$ not meager.
+                    $\emptyset\neq  P \cap \underbrace{\{x : (P \times Q)_x \text{ meager} \}}_{\text{comeager}} \ni x$.
+                    $(P \times Q)_x = Q$ is meager.
+            \end{itemize}
+        \item (ii) $\impliedby$:
+            \begin{itemize}
+                \item $A$ BP, $\{x : A_x \text{ meager}\}$ comeager.
+                \item $A = U \symdif M$.
+                \item Suppose  $A$ not meager $\leadsto$ $U$ not meager
+                    $\leadsto \exists G \times H \subseteq U$ not meager.
+                \item $G$ and $H$ are not meager.
+                \item $\exists x_0 \in G \cap \underbrace{\{x: A_x \text{ meager } \land M_x \text{ meager}\}}_\text{comeager}$.
+                \item $H$ meager, as
+                    \[
+                    H \subseteq U_{x_0} \subseteq A_{x_0} \cup M_{x_0}.
+                    \]
+            \end{itemize}
+    \end{itemize}
+}
 \end{refproof}
+\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}
+}{}
 
 \section{Borel sets} % TODO: fix chapters
 

inputs/lecture_17.tex

@@ -127,7 +127,7 @@ since $X^X$ has these properties.
 
 \begin{lemma}[Ellis–Numakura]
     \yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
-    Every compact semigroup
+    Every non-empty compact semigroup
     contains an \vocab{idempotent} element,
     i.e.~$f$ such that $f^2 = f$.
 \end{lemma}

inputs/lecture_20.tex

@@ -20,9 +20,6 @@
 \end{remark}
 }{}
 
-
-% TODO ANKI-MARKER
-
 We will be studying projections to the first $d$ coordinates,
 i.e.
 \[
@@ -49,6 +46,9 @@ Let $\pi_n\colon X \to  (S^1)^n$ be the projection to the first $n$
 coordinates.
 
 
+% TODO ANKI-MARKER
+
+
 \begin{lemma}
     \label{lem:lec20:1}
     Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$

inputs/lecture_22.tex

@@ -146,9 +146,7 @@ For this we define
             %  TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$,
             % $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$.
         \item Minimality:%
-            \gist{%
-                \footnote{This is not relevant for the exam.}
-
+            \notexaminable{%
                 Let $\langle E_n : n < \omega \rangle$
                 be an enumeration of a countable basis for $\mathbb{K}^I$.
 
@@ -165,11 +163,10 @@ For this we define
                 is dense in $\overline{x} \mapsto f(\overline{x})$.
                 Since the flow is distal, it suffices to show
                 that one orbit is dense (cf.~\yaref{thm:distalflowpartition}).
-            }{ Not relevant for the exam.}
+            }
 
         \item The order of the flow is $\eta$:%
-            \gist{%
-                \footnote{This is not relevant for the exam.}
+            \notexaminable{%
                 Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
                 Consider the flows we get from $(f_i)_{i < j}$
                 resp.~$(f_i)_{i \le j}$
@@ -193,6 +190,6 @@ For this we define
                 \end{IEEEeqnarray*}
                 Beleznay and Foreman show that this is open and dense.%
                 % TODO similarities to the lemma used today
-            }{ Not relevant for the exam.}
+            }
     \end{itemize}
 \end{proof}

inputs/lecture_23.tex

@@ -37,10 +37,9 @@ Let $I$ be a linear order
             S & \coloneqq  & \{ x \in \LO(\N) :& x \text{ has a least element},\\
               &&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
         \end{IEEEeqnarray*}
-        \todo{Exercise sheet 12}
-        $S$ is Borel.
+        $S$ is Borel.\footnote{cf.~\yaref{s12e1}}
 
-        We will % TODO ? 
+        We will
         construct a reduction
         \begin{IEEEeqnarray*}{rCl}
             M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\

inputs/lecture_24.tex

@@ -1,10 +1,10 @@
 \lecture{24}{2024-01-23}{Combinatorics!}
 
+% ANKI 2
+
 
 \subsection{Applications to Combinatorics} % Ramsey Theory}
 
-% TODO Define Ultrafilter
-
 \begin{definition}
     An \vocab{ultrafilter} on $\N$ (or any other set)
     is a family $\cU \subseteq \cP(\N)$
@@ -44,6 +44,7 @@
     for $\{ n \in \N : \phi(n)\} \in \cU$.
     We say that $\phi(n)$ holds for  \vocab{$\cU$-almost all}  $n$.
 \end{notation}
+\gist{%
 \begin{observe}
     Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
 
@@ -53,6 +54,7 @@
         \item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$.
     \end{enumerate}
 \end{observe}
+}{}
 \begin{lemma}
     \label{lem:ultrafilterlimit}
     Let $X $ be a compact Hausdorff space.
@@ -70,7 +72,10 @@
 \begin{notation}
     In this case we write $x = \ulim{\cU}_n x_n$.
 \end{notation}
-\begin{refproof}{lem:ultrafilterlimit}[sketch]
+\begin{refproof}{lem:ultrafilterlimit}\footnote{The proof from the lecture only works
+    for metric spaces.}
+\gist{
+    For metric spaces: 
     Whenever we write $X = Y \cup Z$
     we have $(\cU n) x_n \in Y$
     or $(\cU n) x_n \in Z$.
@@ -85,8 +90,13 @@
     $C \in \cP_{n+1} \implies \exists  C \subseteq D \in \cP_{n}$
     and
      $C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$.
-    It is clear that we can do this for metric spaces,
-    but such partition can be found for compact Hausdorff spaces as well.
+    It is clear that we can do this for metric spaces.
+
+}{}
+    See \yaref{thm:uflimit} for the full proof.
+    See
+    \yaref{fact:compactiffufconv} and
+    \yaref{fact:hdifffilterlimit} for a more general statement.
 \end{refproof}
 
 Let $\beta \N$ be the Čech-Stone compactification of $\N$,
@@ -120,15 +130,14 @@ This gives $+ \colon \beta\N \times  \beta\N \to  \beta\N$.
 
 This is not commutative,
 but associative and $a \mapsto a + b$ is continuous
-for a fixed  $b$.
-This is called a left compact topological semigroup.
-
+for a fixed  $b$,
+i.e.~it is a left compact topological semigroup.
 
 
 
 Let $X$ be a compact Hausdorff space
 and let $T \colon  X \to  X$ be continuous.%
-\footnote{Note that this need not be a homeomorphism, i.e.~we only get a $\N$-action
+\footnote{Note that this may not be a homeomorphism, i.e.~we only get a $\N$-action
     but not a $\Z$-action.}
 
 For any $\cU \in \beta\N$, we define $T^{\cU}$ by
@@ -157,7 +166,6 @@ is not necessarily continuous.
     \[
     \forall n.~\exists k < M.~ T^{n+k}(x) \in G.
     \]
-
 \end{definition}
 \begin{fact}
     Let $\cU, \cV \in \beta\N$

inputs/lecture_25.tex

@@ -7,15 +7,17 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
             where a basis consist of sets
             $V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.
 
+            \gist{%
             (For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
             and $\beta\N = V_\N$.)
+            }{}
 
         \item Note also that for $A, B \subseteq \N$,
             $V_{A \cup B} = V_A \cup V_B$,
             $V_{A^c} = \beta\N \setminus V_A$.
     \end{itemize}
 \end{fact}
-
+\gist{%
 \begin{observe}
     \label{ob:bNclopenbasis}
     Note that the basis is clopen. In particular
@@ -25,6 +27,7 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
     If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
     so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
 \end{observe}
+}{}
 
 \begin{fact}
     \label{fact:bNhd}
@@ -54,12 +57,14 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
     $\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.
 
     We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
+    \gist{%
     Replacing each $F_i$ by $V_{A_j^i}$ such
     that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
     (cf.~\yaref{ob:bNclopenbasis})
     we may assume that $F_i$ is of the form $V_{A_i}$.
     We get $\{F_i = V_{A_i} : i \in I\}$
     with the finite intersection property.
+    }{Wlog.~$F_i = V_{A_i}$.}
     Hence
     $\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
     has the finite intersection property.
@@ -78,11 +83,10 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
         \item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
         \item $\N \subseteq  \beta\N$ is dense.
     \end{itemize}
-    \todo{Easy exercise}
-    % TODO write down (exercise)
 \end{fact}
 
 \begin{theorem}
+    \label{thm:uflimit}
     For every compact Hausdorff space $X$,
     a sequence $(x_n)$ in $X$,
     and $\cU \in \beta\N$,
@@ -132,6 +136,11 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
     % TODO general fact: continuous functions agreeing on a dense set
     % agree everywhere (fact section)
 \end{proof}
+\begin{trivial}+
+    $\beta$ is a functor from the category of topological
+    spaces to the category of compact Hausdorff spaces.
+    It is left adjoint to the inclusion functor.
+\end{trivial}
 
 % RECAP
 \gist{%
@@ -216,13 +225,12 @@ to obtain
             Take $x_2 > x_1$ that satisfies this.
         \item Suppose we have chosen $\langle x_i : i < n \rangle$.
             Since $\cU$ is idempotent, we have
-            \[
-                (\cU n)[
-                    n \in P
-                    \land (\cU_k) n + k \in  P
-                    \land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)
-                    \land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
-            \]
+            \begin{IEEEeqnarray*}{rCl}
+                (\cU n)&& n \in P\\
+                       &\land& (\cU_k) n + k \in  P\\
+                       &\land& \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)\\
+                       &\land& (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
+            \end{IEEEeqnarray*}
             Chose $x_n > x_{n-1}$ that satisfies this.
     \end{itemize}
     Set $H \coloneqq \{x_i : i < \omega\}$.
@@ -231,6 +239,3 @@ to obtain
 
 Next time we'll see another proof of this theorem.
 
-
-
-

inputs/tutorial_02.tex

@@ -3,12 +3,36 @@
 % Points: 15 / 16
 
 \nr 1
-\todo{handwritten solution}
+Let $(X,d)$ be a metric space and $\emptyset \neq A \subseteq X$.
+Let $d(x,A) \coloneqq  \inf(d(x,a) : a \in A\}$.
+
+\begin{itemize}
+    \item $d(-,A)$ is uniformly continuous:
+
+        Clearly $|d(x,A) - d(y,A)| \le  d(x,y)$.
+        \todo{Add details}
+    \item  $d(x,A) = 0 \iff x \in  \overline{A}$.
+
+        $d(x,A) = 0$ iff there is a sequence in $A$
+        converging towards $x$ iff $x \in \overline{A}$.
+\end{itemize}
+
 
 \nr 2
 
+Let $X$ be a discrete space.
+For $f,g \in X^{\N}$ define
+\[
+d(f,g) \coloneqq \begin{cases}
+    (1 + \min \{n: f(n) \neq g(n)\})^{-1} &: f \neq g,\\
+    0 &: f= g.
+\end{cases}
+\]
+
+
 \begin{enumerate}[(a)]
-    \item $d$ is an ultrametric:
+    \item $d$ is an \vocab{ultrametric},
+        i.e.~$d(f,g) \le \max \{d(f,h), d(g,h)\}$ for all $f,g,h \in  X^{\N}$ :
 
         Let $f,g,h \in X^{\N}$.
 
@@ -70,10 +94,15 @@
 
 \nr 3
 
+Consider $\N$ as a discrete space and $\N^{\N}$ with the product topology.
+Let
+\[
+S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}.
+\]
 \begin{enumerate}[(a)]
     \item $S_{\infty}$ is a Polish space:
 
-        From (2) we know that $\N^{\N}$ is Polish.
+        From \yaref{s1e2} we know that $\N^{\N}$ is Polish.
         Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
         with respect to $\N^\N$.
 
@@ -111,69 +140,9 @@
         Clearly there cannot exist a finite subcover
         as $B$ is the disjoint union of the $B_j$.
 
-         % TODO Think about this
 \end{enumerate}
 
 \nr 4
-
-% (uniform metric)
-% 
-% \begin{enumerate}[(a)]
-%     \item $d_u$ is a metric on $\cC(X,Y)$:
-% 
-%         It is clear that $d_u(f,f) = 0$.
-% 
-%         Let  $f \neq g$. Then there exists $x \in X$ with
-%         $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$.
-% 
-%         Since $d$ is symmetric, so is $d_u$.
-% 
-%         Let $f,g,h \in \cC(X,Y)$.
-%         Take some $\epsilon > 0$
-%         choose $x_1, x_2 \in X$
-%         with $d_u(f,g) \le   d(f(x_1), g(x_1)) + \epsilon$,
-%         $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$.
-% 
-%         Then for all $x \in X$
-%         \begin{IEEEeqnarray*}{rCl}
-%             d(f(x), h(x)) &\le &
-%             d(f(x), g(x)) + d(g(x), h(x))\\
-%                           &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\
-%                           &\le & d_u(f,g) + d_u(g,h) - 2\epsilon.
-%         \end{IEEEeqnarray*}
-%         Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$.
-%         Taking $\epsilon \to 0$ yields the triangle inequality.
-% 
-%     \item $\cC(X,Y)$ is a Polish space:
-%         \todo{handwritten solution}
-% 
-%         \begin{itemize}
-%             \item $d_u$ is a complete metric:
-% 
-%                 Let $(f_n)_n$ be a Cauchy series with respect to $d_u$.
-% 
-%                 Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect
-%                 to  $d$ for every $x$.
-%                 Hence there exists a pointwise limit $f$ of the $f_n$.
-%                 We need to show that $f$ is continuous.
-% 
-%                 %\todo{something something uniform convergence theorem}
-% 
-%             \item $(\cC(X,Y), d_u)$ is separable:
-% 
-%                 Since $Y$ is separable, there exists a countable
-%                 dense subset $S \subseteq Y$.
-% 
-%                 Consider $\cC(X,S) \subseteq  \cC(X,Y)$.
-%                 Take some $f \in \cC(X,Y)$.
-%                 Since $X$ is compact,
-% 
-% 
-%                 % TODO
-% 
-%         \end{itemize}
-% \end{enumerate}
-
 \begin{fact}
     Let $X $ be a compact Hausdorff space.
     Then the following are equivalent:
@@ -205,7 +174,7 @@
 Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
 and $Y $ Polish.
 Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
-Consider the metric $d_u(f,g) \coloneqq  \sup_{x \in X} |d(f(x), g(x))|$.
+Consider the \vocab{uniform metric} $d_u(f,g) \coloneqq  \sup_{x \in X} |d(f(x), g(x))|$.
 Clearly $d_u$ is a metric.
 
 \begin{claim}
@@ -243,7 +212,7 @@ Clearly $d_u$ is a metric.
     for each $y \in X_m$.
     Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
     Indeed if $f \in \cC(X,Y)$ and $\eta > 0$,
-    we finde $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
+    we find $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
     since $f$ is uniformly continuous.
     Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$.
     We have $d_u(f,g) \le \eta$,

inputs/tutorial_03.tex

@@ -12,6 +12,14 @@
 
 \nr 1
 
+Let $X$ be a Polish space.
+Then there exists an injection $f\colon  X \to 2^\omega$
+such that for each $n < \omega$,
+the set $f^{-1}(\{(y_n) \in 2^\omega : y_n = 1\})$
+is open.
+Moreover if $V \subseteq 2^{ \omega}$ is closed,
+then $f^{-1}(V)$ is $G_\delta$.
+
 Let $(U_i)_{i < \omega}$ be a countable base of $X$.
 Define
 \begin{IEEEeqnarray*}{rCl}
@@ -19,6 +27,7 @@ Define
     x &\longmapsto & (x_i)_{i < \omega}
 \end{IEEEeqnarray*}
 where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
+\gist{
 Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
 is open.
 We have that $f$ is injective since $X$ is T1.
@@ -51,17 +60,21 @@ Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$
 is finite, we get that
 $f^{-1}(2^{\omega} \setminus  ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
 is $G_\delta$ as a finite union of $G_{\delta}$ sets.
+}{}
 
 
 
 \nr 2
+Let $X$ be a Polish space. Then $X$ is homeomorphic to a closed subspace of $\R^{ \omega}$ :
 \todo{handwritten solution}
-(b)
-Let $f(x^{(i)})$ be a sequence in $f(X)$.
-Suppose that $f(x^{(i)}) \to y$.
-We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
-Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
-Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
+% \begin{itemize}
+%     \item 
+%         Let $f(x^{(i)})$ be a sequence in $f(X)$.
+%     Suppose that $f(x^{(i)}) \to y$.
+%     We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
+%     Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
+%     Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
+% \end{itemize}
 
 
 \nr 3
@@ -130,6 +143,13 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
 \end{proof}
 
 \nr 4
+
+Define
+\begin{IEEEeqnarray*}{rCl}
+    f\colon \omega^{\omega} &\longrightarrow & 2^\omega \\
+    (x_n)&\longmapsto & 0^{x_0} 1 0^{x_1} 1 \ldots.
+\end{IEEEeqnarray*}
+
 \begin{enumerate}[(1)]
     \item $f$ is a topological embedding:
         Consider a basic open set

inputs/tutorial_04.tex

@@ -8,7 +8,7 @@ Let $A \neq \emptyset$ be discrete.
 For $D \subseteq  A^{\omega}$,
 let
 \[
-    T_D \coloneqq  \{x\defon{n} \in  A^{<\omega} | x \in D, n \in \N\}..
+    T_D \coloneqq  \{x\defon{n} \in  A^{<\omega} | x \in D, n \in \N\}.
 \]
 \begin{enumerate}[(a)]
     \item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:

inputs/tutorial_11.tex

@@ -63,7 +63,7 @@ Flows are always on non-empty spaces $X$.
 \begin{proof}
     (i) $\implies$ (ii):
     Let $(Y,T)$ be a subflow of $(X,T)$.
-    take  $y \in Y$. Then $Ty$ is dense in mKX.
+    take  $y \in Y$. Then $Ty$ is dense in $X$.
     But $Ty \subseteq Y$, so $Y$ is dense in $X$.
     Since $Y$ is closed, we get $Y = X$.
 

inputs/tutorial_14.tex

@@ -84,11 +84,12 @@ with parameter $\alpha \in \R$, $1 \cdot x \coloneqq  x + \alpha$.
 \nr 4
 
 % Examinable!
+% TODO THINK!
 
+\gist{%
 % RECAP
-Let $X$ be a metrizable topological space.
-
-Let $K(X) \coloneqq  \{ K \subseteq  X : \text{ compact}\}$.
+Let $X$ be a metrizable topological space
+and let $K(X) \coloneqq  \{ K \subseteq  X : K \text{ compact}\}$.
 
 The Vietoris topology has a basis given by
 $\{K \subseteq U\}$, $U$ open (type 1)
@@ -103,19 +104,21 @@ $\max_{a \in A} d(a,B)$.
 On previous sheets, we checked that $d_H$ is a metric.
 If $X$ is separable, then so is $K(X)$.
 % END RECAP
+}{}
 
 \begin{fact}
+    \label{fact:s12e4}
 Let $(X,d)$ be a complete metric space.
 Then so is  $(K(X), d_H)$.
 \end{fact}
-\begin{proof}
+\begin{refproof}{fact:s12e4}
     We need to show that  $(K(X), d_H)$ is complete.
 
     Let $(K_n)_{ n< \omega}$ be Cauchy in $(K(X), d_H)$.
     Wlog.~$K_n \neq \emptyset$ for all $n$.
 
     Let $K = \{ x \in X : \forall  x \in U \overset{\text{open}}{\subseteq} X.~
-    \text{ $X$ intersects $K_n$ for infinitely many $n$}\}$.
+    \text{ $U \cap K_n \neq \emptyset$ for infinitely many $n$}\}$.
 
     Equivalently,
     $K = \{x : x \text{ is a cluster point of some subsequence $(x_n)$  with $x_n \in K_n$ for all $K_n$}\}$.
@@ -123,12 +126,12 @@ Then so is  $(K(X), d_H)$.
     (A cluster point is a limit of some subsequence).
 
     \begin{claim}
+        \label{fact:s12e4:c1}
         $K_n \to K$.
     \end{claim}
-    \begin{subproof}
+    \begin{refproof}{fact:s12e4:c1}
     Note that $K$ is closed (the complement is open).
 
-
     \begin{claim}
         $K \neq \emptyset$.
     \end{claim}
@@ -159,7 +162,7 @@ Then so is  $(K(X), d_H)$.
         space, it is complete.
 
         So it suffices to show that $K$ is totally bounded.
-        Let $\epsilon > 0$
+        Let $\epsilon > 0$.
         Take $N$ such that $d_H(K_i,K_j) < \epsilon$
         for all $i,j \ge N$.
 
@@ -200,9 +203,8 @@ Then so is  $(K(X), d_H)$.
     To do this, construct a sequence of $y_{n_i} \in K_{n_i}$
     starting with $y$ such that $d(y_{n_i}, y_{n_{i+1}}) < \frac{\epsilon}{2^{i+2}}$.
     (same trick as before).
-    \end{subproof}
-
-\end{proof}
+    \end{refproof}
+\end{refproof}
 
 \begin{fact}
     If $X$ is compact metrisable,
@@ -223,9 +225,3 @@ Then so is  $(K(X), d_H)$.
 
 
 % TODO complete and totally bounded Sutherland metric and topological spaces
-
-
-
-
-
-

inputs/tutorial_15.tex

@@ -2,7 +2,7 @@
 \tutorial{15}{2024-01-31}{Additions}
 
 The following is not relevant for the exam,
-but gives a more general picture.
+but aims to give a more general picture.
 
 Let $X$ be a topological space
 and let $\cF$ be a filter on $ X$.
@@ -12,6 +12,7 @@ all sets containing an open neighbourhood of $x$,
 is contained in  $\cF$.
 
 \begin{fact}
+    \label{fact:hdifffilterlimit}
     $X$ is Hausdorff iff every filter has at most one limit point.
 \end{fact}
 \begin{proof}
@@ -21,6 +22,7 @@ is contained in  $\cF$.
 \end{proof}
 
 \begin{fact}
+    \label{fact:compactiffufconv}
     $X$ is (quasi-) compact
     iff every ultrafilter converges.
 \end{fact}
@@ -29,7 +31,7 @@ is contained in  $\cF$.
     Let $\cU$ be an ultrafilter.
     Consider the family $\cV = \{\overline{A} : A \in \cU\}$
     of closed sets.
-    By the FIP we geht that there exist
+    By the FIP we get that there exist
     $c \in X$ such that $c \in \overline{A}$ for all $A \in \cU$.
     Let $N$ be an open neighbourhood of $c$.
     If $N^c \in \cU$, then $c \in N^c \lightning$
@@ -69,17 +71,19 @@ so is $f(\cB)$.
 \end{fact}
 \begin{proof}
     Consider $(f,g)^{-1}(\Delta) \supseteq A$.
+    The RHS is a dense closed set, i.e.~the entire space.
 \end{proof}
 
-We can uniquely extend $f\colon  X \to  Y$ continuous
+We can uniquely extend a continuous $f\colon  X \to  Y$
 to a continuous $\overline{f}\colon  \beta X \to  Y$
 by setting $\overline{f}(\cU) \coloneqq  \lim_\cU f$.
 
-Let $V$ be an open neighbourhood of $Y$ in $\overline{f}\left( U) \right) $.
-Consider $f^{-1}(V)$.
-Consider the basic open set
-\[
-\{\cF \in  \beta\N  : \cF \ni f^{-1}(V)\}.
-\]
+% Let $V$ be an open neighbourhood of $y \in \overline{f}\left( U \right)$.
+% Consider $f^{-1}(V)$.
+% Then
+% \[
+% \{\cF \in  \beta\N  : \cF \ni f^{-1}(V)\}
+% \]
+% is a basic open set.
 
 \todo{I missed the last 5 minutes}