additional tutorial

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Josia Pietsch 2024-02-02 01:48:47 +01:00
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9 changed files with 125 additions and 12 deletions

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

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

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

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@ -170,7 +170,7 @@ Recall:
Let $(T,X)$ and $(T,Y)$ be flows. Let $(T,X)$ and $(T,Y)$ be flows.
A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$ A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$
is a continuous surjection $X \twoheadrightarrow Y$ is a continuous surjection $X \twoheadrightarrow Y$
commuting with the group action, that is $T$-equivariant,
i.e.~$\forall t \in T, x \in X.~\pi(t\cdot x) = t\cdot \pi(x)$. i.e.~$\forall t \in T, x \in X.~\pi(t\cdot x) = t\cdot \pi(x)$.
If such a factor map exists, If such a factor map exists,
we also say that $(T,Y)$ is a \vocab{factor} we also say that $(T,Y)$ is a \vocab{factor}
@ -257,7 +257,9 @@ Recall:
Let $\Sigma = \{(X_i, T) : i \in I\} $ Let $\Sigma = \{(X_i, T) : i \in I\} $
be a collection of factors of $(X,T)$. % TODO State precise definition of a factor be a collection of factors of $(X,T)$. % TODO State precise definition of a factor
Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map. Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
Then $(X, T)$ is the \vocab{limit} of $\Sigma$ Then $(X, T)$ is a \vocab{limit}%
\footnote{This does not seem to be a limit in the category theory sense.}
of $\Sigma$
iff iff
\[ \[
\forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2). \forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).

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

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

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@ -11,6 +11,26 @@
\nr 3 \nr 3
% somewhat examinable (for 1.0) % 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 \nr 4

87
inputs/tutorial_15.tex Normal file
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\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
\[
\{\cF \in \beta\N : \cF \ni f^{-1}(V)\}.
\]

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\input{inputs/tutorial_12b} \input{inputs/tutorial_12b}
\input{inputs/tutorial_12} \input{inputs/tutorial_12}
\input{inputs/tutorial_14} \input{inputs/tutorial_14}
\input{inputs/tutorial_15}
\section{Facts} \section{Facts}
\input{inputs/facts} \input{inputs/facts}