tutorial 09
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Josia Pietsch 2023-12-12 13:31:59 +01:00
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@ -12,6 +12,7 @@
\end{definition}
\begin{proposition}
\label{prop:ifs11}
$\IF \in \Sigma^1_1(\Tr)$.
\end{proposition}
\begin{proof}

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@ -197,16 +197,9 @@ However if we pick $y \in Y$, $Ty$ might not be dense.
% so $Gy \subseteq Gx$.
% Since $y = g_0 x \implies x = g_0^{-1}y$, we also have $x \in Gy$,
% hence $Gx \subseteq Gy$
% TODO: WHY?
\end{proof}
\begin{corollary}
If $(X,T)$ is distal and minimal,
then $E(X,T) \acts X$ is transitive.
\end{corollary}

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@ -125,7 +125,7 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
% Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
% with $Y \subseteq G \subseteq \overline{Y}$.
% $\tilde{f}$ and $\id_G$ agree on $Y$.
% $Y$ is dense in $G$ and the codomain of $f$ is ltd.\todo{????}
% $Y$ is dense in $G$ and the codomain of $f$ is Hausdorff.\todo{????}
% So $f = \id_G$, i.e.~$G = Y$.
\end{proof}

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@ -109,8 +109,6 @@ for some $B_i \in \cB(Y_i)$.
\end{itemize}
\nr 3
\todo{Wait for mail}
\begin{lemma}
Let $X$ be a second-countable topological space.
Then every base of $X$ contains a countable subset which
@ -134,6 +132,9 @@ for some $B_i \in \cB(Y_i)$.
such that $m \in M_{f(m)}$.
Then $\bigcup_{m \in M} B_{f(m)} = C_n$.
\end{proof}
\begin{remark}
We don't actually need this.
\end{remark}
\begin{itemize}
\item We use the same construction as in exercise 2 (a)
@ -188,31 +189,71 @@ for some $B_i \in \cB(Y_i)$.
\]
is countable as a countable union of countable sets $\lightning$.
\item Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
as in the first part.
Clearly $f$ is also continuous with respect to the new topology,
so we may assume that $X$ is zero dimensional.
Let $W \subseteq X$ be such that $f\defon{W}$ is injective
and $f(W) = f(X)$ (this exists by the axiom of choice).
Since $f(X)$ is uncountable, so is $W$.
By the second point, there exist disjoint clopen sets
$U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
are uncountable.
Inductively construct $U_s$ for $s \in 2^{<\omega}$
as follows:
Suppose that $U_{s}$ has already been chosen.
Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
be disjoint clopen such that $U_{s\concat 1} \cap W$
and $U_{s\concat 0} \cap W$ are uncountable.
Such sets exist, since $ U_s \cap W$ is uncountable
and $U_s$ is a zero dimensional space with the subspace topology.
And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
in $U_s$ iff it is clopen in $X$.
Other proof (without using the existence of a countable clopen
basis):
We can cover $X$ by countably many clopen sets of diameter
$< \frac{1}{n}$:
Cover $X$ with open balls of diameter $< \frac{1}{n}$.
Write each open ball as a union of clopen sets.
That gives us a cover by clopen sets of diameter $< \frac{1}{n}$.
As $X$ is Lindelöf, there exists a countable subcover.
Then continue as in the first proof.
\item Note that this step does not help us to prove the statement.
It was an error on the exercise sheet.
% Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
% as in the first part.
% Clearly $f$ is also continuous with respect to the new topology,
% so we may assume that $X$ is zero dimensional.
%
% Let $W \subseteq X$ be such that $f\defon{W}$ is injective
% and $f(W) = f(X)$ (this exists by the axiom of choice).
% Since $f(X)$ is uncountable, so is $W$.
% By the second point, there exist disjoint clopen sets
% $U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
% are uncountable.
% Inductively construct $U_s$ for $s \in 2^{<\omega}$
% as follows:
% Suppose that $U_{s}$ has already been chosen.
% Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
% be disjoint clopen such that $U_{s\concat 1} \cap W$
% and $U_{s\concat 0} \cap W$ are uncountable.
% Such sets exist, since $ U_s \cap W$ is uncountable
% and $U_s$ is a zero dimensional space with the subspace topology.
% And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
% in $U_s$ iff it is clopen in $X$.
Clearly this defines a Cantor scheme.
\item \todo{TODO}
\item Let $Y$ be a Polish space and $A \subseteq Y$ analytic
and uncountable.
Expand the topology on $Y$ so that $Y$ is zero dimensional
and $A$ is still analytic.
Then there exists a Polish space $X$
and a continuous function $f\colon X \to Y$
such that $f(X) = A$.
$A$ is uncountable, so by (2)
there exists non-empty disjoint clopen $V_0$, $V_1$
such that $V_0 \cap A$ and $V_1 \cap A$
are uncountable.
Let $W_0 = f^{-1}(V_0 \cap A)$ and $W_1 = f^{-1}(V_1 \cap A)$.
$ W_0$ and $W_1$ are clopen and disjoint.
We can cover $W_0$ with countably many open sets
of diameter $\le \frac{1}{n}$
and similarly for $W_1$.
Then pick open sets such that there image is uncountable.
Repeating this construction we get a Cantor scheme on $X$.
So we get $2^{\N} \overset{s}{\hookrightarrow} X$
and by construction of the cantor scheme,
we get that $f \circ s$ is injective and continuous.
\end{itemize}
\nr 4

159
inputs/tutorial_09.tex Normal file
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@ -0,0 +1,159 @@
\tutorial{09}{2023-12-12}{}
\begin{fact}
Let $X,Y$ be topological spaces
$X$ (quasi-)compact and
$Y$ Hausdorff.
Let $f\colon X\to Y$ be a continuous bijection.
Then $f$ is a homeomorphism.
\end{fact}
\begin{proof}
Compact subsets of Hausdorff spaces are closed.
\end{proof}
\subsection{Sheet 8}
Material on topological dynamic
\begin{itemize}
\item Terence Tao's notes on ergodic theory 254 A.
\item Furstenberg (uses very different notation!).
\end{itemize}
\nr 1
\begin{remark}
$\Sigma^1_1$-complete sets are in some sense the ``worst''
$\Sigma^1_1$-sets:
Deciding whether an element is contained in the
$\Sigma^1_1$-complete set
is at least as ``hard'' as as for any $\Sigma^1_1$ set.
In particular, $\Sigma^1_1$-complete sets
are not Borel.
\end{remark}
Similarly as in \yaref{prop:ifs11} it can be shown that $L \in \Sigma^1_1(X)$:
Consider $\{(x, \beta) \in X \times \cN : \forall n.~x_{\beta_n} | x_{\beta_{n+1}}\}$.
This is closed in $X \times \cN$, since it is a countable intersection of clopen
sets and $L = \proj_X(D)$.
Since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete,
it suffices to find a Borel map $f\colon \Tr \to X$
such that $x \in \IF \iff f(x) \in L$.
Let $\phi\colon \omega^2 + \omega \to \omega$ be bijective
and let $p_i$ denote the $i$-th prime.
Define
\begin{IEEEeqnarray*}{rCl}
\psi\colon \omega^{<\omega} &\longrightarrow & \omega \setminus \{0\} \\
(s_0, s_1, \ldots, s_{n-1})&\longmapsto & \prod_{i < n} p_{\phi(\omega \cdot i + s_i)}.
\end{IEEEeqnarray*}
Note that $\psi$ is injective and that $s \in \omega^{<\omega}$ is an initial segment
of $t \in \omega^{<\omega}$ iff $\psi(s) | \psi(t)$.
Let
\begin{IEEEeqnarray*}{rCl}
f' \colon \Tr &\longrightarrow & \cP(\omega \setminus \{0\}) \\
T &\longmapsto & \{\phi(s) : s \in T\}.
\end{IEEEeqnarray*}
We can turn this into a function
$f\colon \Tr \to (\omega \setminus \{0\})^{\omega}$
by mapping a subset of $\omega \setminus \{0\}$
to the unique strictly increasing sequence whose elements are from
that subset
(appending $\phi(\omega^2 + n), n \in \omega$, if the subset was finite).
Note that $T \in \IF \iff f(T) \in L$.
Furthermore $f$ is Borel, since fixing
a finite initial sequence (i.e.~a basic open set of $(\omega \setminus \{0\})^{\omega}$)
amounts to a finite number of conditions on the preimage.
\nr 2
\todo{handwritten}
% Aron
%
% \begin{enumerate}[1.]
% \item This is trivial $\sup \sup$.
% \item Clearly there are trees of rank $n$ for all $n < \omega$.
% Glue them together.
% \[
% \{(0,0,i,\underbrace{0,\ldots,0}_{i \text{~times}}) | i < \omega\}.
% \]
% \item Map infinite branches. $\sup$ the $\le $.
% \item Induction on $\rho(S)$.
% Cofinal subsequences bla bla.
% \end{enumerate}
\nr 3
\begin{itemize}
\item $LO(\N) \overset{\text{closed}}{\subseteq} 2^{\N\times \N}$:
We have $< \in LO(\N)$ iff
\begin{itemize}
\item $\forall x,y.~ (x \neq y \implies (x< y \lor x > y))$,
\item $\forall x.~(x \not < x)$,
\item $\forall x,y,x.~(x < y < z \implies x < z)$.
\end{itemize}
Write this with $\bigcap$,
i.e.
\begin{IEEEeqnarray*}{rCl}
LO(\N) &=& \bigcap_{n \in \N} \{R: (n,n) \not\in R\}\\
&& \cap \bigcap_{m < n \in \N} (\{R: (n,m) \in R\} \cup \{R: (m,n) \in R\})\\
&&\cap \bigcap_{a,b,c \in \N} (\{R: (a,b) \in R \land (b,c) \in R \implies (a,c) \in R\}.
\end{IEEEeqnarray*}
This is closed as an intersection of clopen sets.
\item We apply \yaref{thm:borel} (iv).
Let $\cF \subseteq LO(\N) \times \cN$ be
such that the $\cN$-coordinate encodes a strictly
decreasing sequence, i.e.~
\[(R, s) \in \cF :\iff \forall n \in \N.~(s(n+1), s(n)) \in R.\]
We have that
\[
\cF = \bigcap_{n \in \N} \{(R,s) \in LO(\N)\times \cN : (s(n+1), s(n)) \in R\}
\]
is closed as an intersection of clopen sets.
Clearly $\pr_{LO(\N)}(\cF)$ is the complement
of $WO(\N)$, hence $WO(\N)$ is coanalytic.
\end{itemize}
\nr 4
\begin{remark}
In the lecture we only look at metrizable flows,
so the definitions from the exercise sheet and from
the lecture don't agree.
\end{remark}
\begin{itemize}
\item Consider
\begin{IEEEeqnarray*}{rCl}
\Phi\colon \Z\text{-flows on } X &\longrightarrow & \Homeo(X) \\
(\alpha\colon \Z\times X \to X) &\longmapsto & \alpha(1, \cdot)\\
\begin{pmatrix*}[l]
\Z\times X &\longrightarrow & X \\
(z,x) &\longmapsto & \beta^{z}(x)
\end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X).
\end{IEEEeqnarray*}
Clearly this has the desired properties.
\item We have
\begin{IEEEeqnarray*}{Cl}
& \Z \circlearrowright X \text{ has a dense orbit}\\
\iff& \exists x \in X.~ \overline{\Z\cdot x} = X\\
\iff& \exists x \in X.~\forall U\overset{\text{open}}{\subseteq} X.~\exists z \in \Z.~
z \cdot x \in U\\
\iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~
\exists z \in \Z.~f^z(x) \in U.
\end{IEEEeqnarray*}
\item Not solved.
\item It suffices to check the condition from part (b)
for open sets $U$ of a countable basis
and points $x \in X$ belonging to a countable dense subset.
Replacing quantifiers by unions resp.~intersections
gives that $D$ is Borel.
\end{itemize}

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@ -134,6 +134,9 @@
\DeclareSimpleMathOperator{WO} % well orderings
\DeclareSimpleMathOperator{Homeo}
\DeclareSimpleMathOperator{osc} % oscillation
\newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}

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@ -60,6 +60,7 @@
\input{inputs/tutorial_06}
\input{inputs/tutorial_07}
\input{inputs/tutorial_08}
\input{inputs/tutorial_09}
\section{Facts}
\input{inputs/facts}