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@ -15,7 +15,7 @@
\end{definition}
Note that Polishness is preserved under homeomorphisms,
i.e.~it is really a topological property.
\gist{%
\begin{example}
\begin{itemize}
\item $\R$ is a Polish space,
@ -26,6 +26,7 @@ i.e.~it is really a topological property.
considered as a topological space.
\end{itemize}
\end{example}
}{}
\gist{%
Polish spaces behave very nicely.
We will see that uncountable polish spaces have size $2^{\aleph_0}$. % TODO: mathfrak c for continuum
@ -75,11 +76,13 @@ However the converse of this does not hold.
\item \vocab{Lindelöf} (every open cover has a countable subcover).
\end{itemize}
\end{fact}
\gist{%
\begin{fact}
Compact Hausdorff spaces are \vocab{normal} (T4)
i.e.~two disjoint closed subsets can be separated
by open sets.
\end{fact}
}{}
\begin{fact}
For a metric space, the following are equivalent:
\begin{itemize}
@ -172,7 +175,8 @@ suffices to show that open balls in one metric are unions of open balls in the o
D\left( (x_n), (y_n) \right) \coloneqq
\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
\]
Clearly $D \le 1$.
\gist{%
Clearly $D \le 1$.
It is also clear, that $D$ is a metric.
We need to check that $D$ is complete:
@ -180,6 +184,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
Consider the pointwise limit $(a_n)$.
This exists since $x_n^{(k)}$ is Cauchy for every fixed $n$.
Then $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
}{Clearly $D$ is a complete metric.}
\end{proof}
\begin{definition}[Our favourite Polish spaces]

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@ -59,7 +59,6 @@
define extension, initial segments
and concatenation of a finite sequence with an infinite one.
\end{notation}
}{}
\begin{definition}
A \vocab{tree}
@ -87,6 +86,7 @@
\forall t\in T.\exists s \supsetneq t.~s \in T.
\]
\end{definition}
}{}
\begin{definition}
A \vocab{Cantor scheme}

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@ -139,7 +139,7 @@
\subsection{The hierarchy of Borel sets}
Let $\omega_1$ be the first uncountable ordinal.
\gist{Let $\omega_1$ be the first uncountable ordinal.}{}
For every $d < \omega_1$,
we define by transfinite recursion
classes $\Sigma^0_\alpha$
@ -201,6 +201,7 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
\end{enumerate}
\end{proposition}
\begin{proof}
\gist{%
\begin{enumerate}[(a)]
\item \begin{observe}
\label{ob:sigmasuffices}
@ -215,7 +216,7 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
since $\Delta^0_\eta(X)$ is closed under complements.
Furthermore, it suffices to show $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$,
by \yaref{ob:sigmasuffices}
by the observation
(since $\Sigma^0_\eta(X) \subseteq \Pi^0_\xi(X)$
and $\Delta^0_\xi(X) = \Sigma^0_\xi(X) \cap \Pi^0_\xi(X)$).
@ -246,6 +247,17 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
Then $\bigcup_{n < \omega} A_n \in \Sigma^0_\alpha(X)$.
It is clear that $\cB_0$ is closed under complements.
\end{enumerate}
}{
\begin{enumerate}[(a)]
\item It suffices to show that $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$
for all $1 \le \eta < \xi < \omega_1$.
For $\eta = 1, \xi = 2$ this holds,
since open sets of a metrizable space are $F_\sigma$.
Induction.
\item Let $\cB_0 \coloneqq \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
This is a $\sigma$-algebra containing all open sets.
\end{enumerate}
}
\end{proof}

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@ -6,6 +6,7 @@
% $\fc := 2^{\aleph_0}$
\end{proposition}
\begin{proof}
\gist{%
We use strong induction on $\xi < \omega_1$.
We have $\Sigma^0_1(X) \le \fc$
(for every element of the basis, we can decide
@ -27,6 +28,9 @@
\[
|\cB(X)| \le \omega_1 \cdot \fc = \fc.
\]
}{Use strong induction. $|\Sigma^0_1(X)| \le \fc$, since $X$ is second countable.
\[|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}\]
}
\end{proof}
\begin{proposition}[Closure properties]

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@ -41,6 +41,7 @@ where $X$ is a metrizable, usually second countable space.
similarly for $\Pi^0_\xi(X)$.
\end{theorem}
\begin{proof}
\gist{%
Note that if $\cU$ is $2^{\omega}$ universal for
$\Sigma^0_\xi(X)$, then $(2^{\omega} \times X) \setminus \cU$
is $2^{\omega}$-universal for $\Pi^0_\xi(X)$.
@ -89,6 +90,18 @@ where $X$ is a metrizable, usually second countable space.
Let $A \in \Sigma^0_\xi(X)$.
Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
Furthermore $\cU \in \Sigma^0_{\xi}((2^{\omega \times \omega} \times X)$.
}{
\begin{itemize}
\item Suffices for $\Sigma^0_\xi$ (complement is $\Pi^0_\xi$-universal).
\item $2^\omega$-universal set for $\Sigma^0_1(X)$, since
$X$ is second countable ($(y,x) \in \cU \iff x \in \bigcup_n \{V_n : y_n = 1\}$).
\item Induction: Take $\xi_k \to \xi$,
and $\cU_{\xi_k}$ $\Sigma^0_{\xi_k}$-universal.
Construct $2^{ \omega \times \omega}$-universal:
$(y_{m,n}, x) \in \cU :\iff \exists n.~((y_{m,n}), x) \in \cU_{\xi_n}$.
\end{itemize}
}
\end{proof}
\begin{remark}
Since $2^{\omega}$ embeds
@ -96,6 +109,6 @@ where $X$ is a metrizable, usually second countable space.
% such that the image is closed,
we can replace $2^{\omega}$ by $Y$
in the statement of the theorem.%
\footnote{By definition of the subspace topology
and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
\gist{\footnote{By definition of the subspace topology
and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}}{}
\end{remark}

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@ -108,14 +108,14 @@
\end{theorem}
\begin{proof}
Let $\phi\colon R \to \Ord$
by a $\Pi^1_1$-rank.
Set
be a $\Pi^1_1$-rank.
\gist{Set
\begin{IEEEeqnarray*}{rCl}
(x,n) \in R^\ast &:\iff& (x,n) \in R\\
&&\land \forall m.~(x,n) \le^\ast_\phi (x,m)\\
&&\land \forall m.~\left( (x,n) <^\ast_\phi (x,m) \lor n \le m \right),
\end{IEEEeqnarray*}
i.e.~take the element with minimal rank
i.e.~take}{Take} the element with minimal rank
that has the minimal second coordinate among those elements.
\end{proof}
\gist{
@ -196,8 +196,8 @@ Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
\[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
\begin{itemize}
\item $n < m$ and
\item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.%
\footnote{sic! (there is a typo in the official notes)}
\item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
% \footnote{sic! (there was a typo in the official notes)}
\end{itemize}
\begin{claim}
$\prec^\ast$ is well-founded.

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@ -66,7 +66,7 @@
% \subsection*{Basic Definitions}
% TODO: move to appendix?
\gist{%
Recall:
\begin{definition}+
Let $X$ be a set.
@ -112,6 +112,7 @@ Recall:
g&\longmapsto & (x \mapsto g \cdot x).
\end{IEEEeqnarray*}
\end{remark}
}{}
\begin{definition}+
A group $G$ with a topology
@ -185,13 +186,13 @@ Recall:
a homeomorphism $X \leftrightarrow Y$
commuting with the group action.
\end{definition}
\gist{%
\begin{warning}+
What is called ``factor'' here is called ``subflow''
by Furstenberg.
\end{warning}
\begin{example}
Recall that $S_1 = \{z \in \C : |z| = 1\}$.
Let $X = S_1$, $T = S_1$
@ -200,6 +201,7 @@ Recall:
and $\alpha + \beta$ denotes the addition of \emph{angles},
i.e.~$\alpha \cdot \beta$ in complex numbers.}
\end{example}
}{}
\begin{definition}
\label{def:isometricextension}
@ -237,6 +239,7 @@ Recall:
An isometric extension of a distal flow is distal.
\end{proposition}
\begin{proof}
\gist{%
Let $\pi\colon X\to Y$ be an isometric extension.
Towards a contradiction,
suppose that $x_1,x_2 \in X$ are proximal.
@ -253,6 +256,11 @@ Recall:
we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$.
Therefore $\rho(x_1,x_2) = 0$.
Hence $x_1 = x_2$ $\lightning$.
}{Let $\pi\colon X \to Y$ isometric extension.
Suppose $x_1,x_2 \in X$ is proximal.
Then $\pi(x_1) = \pi(x_2)$.
But there exists a $T$-equivariant metric on the fibers.
}
\end{proof}
\begin{definition}

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@ -149,6 +149,7 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
bet the quotient space.
It is compact, second countable and Hausdorff.
Let $\pi\colon X\to M$ denote the quotient map.
\gist{%
\item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$
is an isometric flow:
\begin{enumerate}
@ -204,6 +205,7 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
(\yaref{thm:usccomeagercont})
this implies that $X \setminus \{x_2\}$ is meager.
But then $X = \{\star\} \lightning$.
}{}
\end{enumerate}
\end{proof}

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@ -173,6 +173,7 @@ More generally we can show:
then $(Y,T)$ is an isometric extension of $(Z_2, T)$.
\end{lemma}
\begin{proof}
\gist{%
% TODO TODO TODO Think about this
For $z_1,z_1' \in Z_1$ with
$w_1(z_1) = w_1(z_1')$ let
@ -185,9 +186,16 @@ More generally we can show:
on the fibers of $Y$ over $Z_2$
and invariant under $T$.
$\sigma$ is a metric,
$\sigma$ is a metric on fibers,
since if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$,
then $\pi_1(y) = \pi_1(y')$ or $y = y'$.
}{%
\begin{itemize}
\item Let $\rho\colon Z_1 \times_W Z_1 \to \R$.
\item Consider $\sigma\colon Y \times_{Z_2} Y \to \R$
given by $\sigma(y,y') \coloneqq \rho(\pi_1(y), \pi_1(y'))$.
\end{itemize}
}
\end{proof}
@ -242,7 +250,7 @@ More generally we can show:
$(X'_\xi, T) = \theta((X_\xi, T)$.
Let $\pi_\xi$ and $\pi'_\xi$ denote the maps from $X$ to $X_\xi$ resp.~$X'_\xi$.
Set
\[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\}.\]
\[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\} \subseteq X \times X'_{\xi+1}.\]
Then
% https://q.uiver.app/#q=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@ -46,9 +46,6 @@ 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 $\pi_n(x) = \pi_n(x')$
@ -66,7 +63,7 @@ coordinates.
where $d$ is the metric on $X$,
$d((x_i), (y_i)) = \max_n \frac{1}{2^n} | x_n - y_n|$.% TODO use multiplicative notation
\end{lemma}
\begin{proof}
\begin{refproof}{lem:lec20:1}
Let
\begin{IEEEeqnarray*}{rCl}
x &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha_{n+1}, \alpha_{n+2},\ldots)\\
@ -74,7 +71,7 @@ coordinates.
\end{IEEEeqnarray*}
We will choose $x_k$ of the form
\[
(\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1} \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
(\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
\]
where $\beta_k$ is such that $\frac{\beta_k}{\pi}$ is irrational
and $|\beta_k| < 2^{-k}$.
@ -89,7 +86,6 @@ coordinates.
$u' = (\xi'_n)_{n \in \N}$,
let $\frac{u}{u'} = (\frac{\xi_n}{\xi'_n})_{n \in \N}$
($X$ is a group).
We are interested in $F(x_k, x') = \inf_m d(\tau^m x_k, \tau^m x')$,
but it is easier to consider the distance between
their quotient and $1$.
@ -97,6 +93,7 @@ coordinates.
\[
w_k \coloneqq \frac{x_k}{x'} = (\underbrace{1,\ldots,1}_{n-1}, e^{\i \beta_k}, \overbrace{\frac{\alpha_{n+1}}{\alpha'_{n+1}}, \frac{\alpha_{n+2}}{\alpha'_{n+2}}, \ldots}^{\mathclap{\text{not interesting}}}).
\]
\gist{%
\begin{claim}
$F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
where $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
@ -121,7 +118,8 @@ coordinates.
&\le & 2^{-n} | e^{\i \beta_k}- 1|\\
&<& 2^{-n-k}.
\end{IEEEeqnarray*}
\end{proof}
}{[some technical details omitted]}
\end{refproof}
\begin{definition}

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@ -1,6 +1,7 @@
\lecture{22}{2024-01-16}{}
\begin{refproof}{thm:21:xnmaxiso}
% TODO TODO TODO
We have the following situation:
% https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0=
\[\begin{tikzcd}
@ -17,7 +18,7 @@
We want to show that this tower is normal,
i.e.~the isometric extensions are maximal isometric extension.
\gist{%
Let $Y$ be a maximal isometric extension of $X_{n-1}$ in $X$
and let $\overline{g} = \pi^n_{n-1} \circ h$. % factor map?
We need to show that $h$ is an isomorphism.
@ -62,6 +63,25 @@
But $x$ and $x'$ don't depend on $k$,
hence $R(x,x') = 0$.
It follows that $\pi'(x) = \pi'(x')$ $\lightning$.
}{
\begin{itemize}
\item $Y$ max.~isometric extension of $X_{n-1}$ in $X$
and $\overline{g} = \pi^n_{n-1} \circ h$.
\item $h$ isomorphism.
Suppose not, then $\exists y_0,y_1 \in X.~\pi'(y_0) \neq \pi'(y_1),
\pi_n(y_0) = \pi_n(y_1) = t$.
\item Apply \yaref{lem:lec20:1} $\leadsto$ sequence $(x_k)$ in $X$,
such that $\pi_{n-1}(x_k) = \pi_{n-1}(y_i)$,
$F(x_k,y_i) \to 0$.
\item $\rho\colon Y \times_{X_{n-1}} Y \to \R$ witnessing isometric.
\item $R(a,b) \coloneqq \rho(\pi'(a), \pi'(b))$ for $a,b \in X$ with
$\overline{g}(\pi'(a)) = \overline{g}(\pi'(b))$.
(defined for any two of $x_k$, $y_0$, $y_1$, $\tau$-equivariant)
\item $F(y_0,x_{k}) \to 0$, so $d(\tau^{m_k} y_0, \tau^{m_k} x_k) \to 0$.
\item $R(y_0,x_k) \to 0$, hence $\underbrace{R(y_0,y_1)}_{\text{no } k} \to 0$ $\lightning$.
\end{itemize}
}
\end{refproof}