inputs/lecture_03.tex

@@ -197,7 +197,8 @@
     Let $X \neq \emptyset$ be a Polish space.
     Then there is a closed subset
     \[
-        D \subseteq \N^\N \mathbin{\text{\reflectbox{$\coloneqq$}}} \cN
+    D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN
+    % TODO correct N for the Baire space?
     \]
     and a continuous bijection from
     $D$ onto $X$ (the inverse does not need to be continuous).
@@ -238,7 +239,7 @@
     \end{enumerate}
 
     \gist{%
-    Suppose we already have $F_s \mathbin{\text{\reflectbox{$\coloneqq$}}} F$.
+    Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
     We need to construct a partition $(F_i)_{i \in \N}$
     of $F$ with $\overline{F_i} \subseteq F$
     and $\diam(F_i) < \epsilon$

inputs/lecture_04.tex

@@ -62,7 +62,7 @@
     We extend $f$ to $g\colon\cN \to X$
     in the following way:
 
-    Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x\defon{n} = s\}$.
+    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 (cf.~\yaref{s3e1})
     \[

inputs/lecture_14.tex

@@ -4,7 +4,6 @@
     If $C$ is coanalytic,
     then there exists a $\Pi^1_1$-rank on $C$.
 \end{theorem}
-% TODO show that WO sse 2^QQ is Pi_1^1 complete
 \begin{proof}
     \gist{%
     Pick a $\Pi^1_1$-complete set.
@@ -26,7 +25,7 @@
     % 	\arrow["\subseteq"', hook, from=2-3, to=1-3]
     % \end{tikzcd}
     Let $X = 2^{\Q} \supseteq \WO$.
-    We have already shown that $\WO$ is $\Pi^1_1$-complete.% TODO REF
+    We have already show that $\WO$ is $\Pi^1_1$-complete.
 
     Set $\phi(x) \coloneqq \otp(x)$
     ($\otp\colon \WO \to \Ord$ denotes the order type).

inputs/lecture_15.tex

@@ -175,7 +175,7 @@ Recall:
     Let $(T,X)$ and $(T,Y)$ be flows.
     A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$
     is a continuous surjection $X \twoheadrightarrow Y$
-    that is $T$-equivariant,
+    commuting with the group action,
     i.e.~$\forall t \in T, x \in X.~\pi(t\cdot x) = t\cdot \pi(x)$.
     If such a factor map exists,
     we also say that $(T,Y)$ is a \vocab{factor}
@@ -258,14 +258,13 @@ Recall:
 \begin{definition}
     Let $\Sigma = \{(X_i, T) : i \in I\} $
     be a collection of factors of $(X,T)$.
-    Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
-    Then $(X, T)$ is a \vocab{limit}%
-    \footnote{This is not a limit in the category theory sense and not uniquely determined.}
-    of $\Sigma$
+    Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor maps.
+    Then $(X, T)$ is the \vocab{limit}  of $\Sigma$
     iff
     \[
     \forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
     \]
+    % TODO think about abstract nonsense
 \end{definition}
 
 \begin{proposition}

inputs/lecture_19.tex

@@ -65,9 +65,9 @@ equicontinuity coincide.
 \end{tikzcd}\]
 \end{theorem}
 \begin{proof}
+    % TODO Think about this
     We want to apply Zorn's lemma.
-    If suffices to show that isometric flows are closed under inverse limits,%
-    \footnote{This seems to be an inverse limit in the category theory sense.}
+    If suffices to show that isometric flows are closed under inverse limits,
     i.e.~if $(Y_\alpha, f_{\alpha,\beta})$,
     $\beta < \alpha \le  \Theta$
     are isometric, then the inverse limit $Y$ is isometric.%

inputs/tutorial_07.tex

@@ -34,9 +34,6 @@
 \end{enumerate}
 
 \nr 2
-
-Recall \yaref{thm:clopenize}:
-
 \begin{fact}
     Let $(X,\tau)$ be a Polish space and
     $A \in \cB(X)$.
@@ -44,10 +41,10 @@ Recall \yaref{thm:clopenize}:
     with the same Borel sets as  $\tau$
     such that $A$ is clopen.
 
+    (Do it for $A$ closed,
+    then show that the sets which work
+    form a $\sigma$-algebra).
 \end{fact}
-(Do it for $A$ closed,
-then show that the sets which work
-form a $\sigma$-algebra).
 
 \begin{enumerate}[(a)]
     \item Let $(X, \tau)$ be Polish.

inputs/tutorial_14.tex

@@ -11,26 +11,6 @@
 
 \nr 3
 % somewhat examinable (for 1.0)
-% TODO
-
-\begin{enumerate}[(a)]
-    \item $(X,T)$ is distal iff it does not have a proximal pair,
-        i.e.~$a\neq b$, $c$ such that $t_n \in T$,
-        $t_na, t_nb \to c$.
-
-        Equivalently,
-        for all $a,b$ there exists an  $\epsilon$,
-        such that for all $t \in T$, $d(ta,tb) > \epsilon$.
-
-
-    \item % TODO (not too hard)
-        % (b)
-        % Let $(X,T)$ be distal with a dense orbit,
-        % then it is distal minimal.
-        % Sheet 8: has dense orbit is Borel
-        % Distal flow decomposes into distal minimal flows.
-\end{enumerate}
-
 
 \nr 4
 

inputs/tutorial_15.tex

@@ -1,87 +0,0 @@
-\tutorial{15}{2024-01-31}{Additions}
-
-\subsection{Additional Tutorial}
-
-The following is not relevant for the exam,
-but gives a more general picture.
-
-Let $ X$ be a topological space.
-Let $\cF$ be a filter on $ X$.
-
-$x \in X$ is a limit point of $\cF$ iff the neighbourhood filter $\cN_x$,
-all sets containing an open neighbourhood of $x$,
-is contained in  $\cF$.
-
-\begin{fact}
-    $X$ is Hausdorff iff every filter has at most one limit point.
-\end{fact}
-\begin{proof}
-    Neighbourhood filters are compatible
-    iff the corresponding points
-    can not be separated by open subsets.
-\end{proof}
-
-\begin{fact}
-    $X$ is (quasi-) compact
-    iff every ultrafilter converges.
-\end{fact}
-\begin{proof}
-    Suppose that $X$ is compact.
-    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
-    $c \in X$ 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 $c \in N^c \lightning$
-    So we get that $N \in \cU$.
-
-    Let $\{V_i : i \in I\} $ be a family of closed sets with the FIP.
-    Consider the filter generated by this family.
-    We extend this to an ultrafilter.
-    The limit of this ultrafilter is contained in all the $V_i$.
-\end{proof}
-
-Let $X,Y$ be topological spaces,
-$\cB$ a filter base on $X$,
-$\cF$ the filter generated by $\cB$
-and
-$f\colon  X \to Y$.
-Then $f(\cB)$ is a filter base on $Y$,
-since $f(\bigcap A_i ) \subseteq  \bigcap f(A_i)$.
-We say that $\lim_\cF f = y$,
-if $f(\cF) \to  y$.
-
-Equivalently $f^{-1}(N) \in \cF$
-for all neighbourhoods $N$ of $y$.
-
-In the lecture we only considered $X = \N$.
-If $\cB$ is the base of an ultrafilter,
-so is $f(\cB)$.
-
-\begin{fact}
-    Let $X$ be a topological space
-    and let $Y$ be Hausdorff.
-    Let $f,g \colon  X \to Y$
-     be continuous.
-     Let $A \subseteq  X$ be dense such that
-     $f\defon{A} = g\defon{A} $.
-     Then $f = g$.
-\end{fact}
-\begin{proof}
-    Consider $(f,g)^{-1}(\Delta) \supseteq A$.
-\end{proof}
-
-We can uniquely extend $f\colon  X \to  Y$ continuous
-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
-\[
-\{\cF \in  \beta\N  : \cF \ni f^{-1}(V)\}.
-\]
-
-
-

logic3.tex

@@ -74,7 +74,6 @@
 \input{inputs/tutorial_12b}
 \input{inputs/tutorial_12}
 \input{inputs/tutorial_14}
-\input{inputs/tutorial_15}
 
 \section{Facts}
 \input{inputs/facts}