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@ -15,7 +15,7 @@
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\end{definition}
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Note that Polishness is preserved under homeomorphisms,
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i.e.~it is really a topological property.
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\gist{%
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\begin{example}
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\begin{itemize}
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\item $\R$ is a Polish space,
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@ -26,6 +26,7 @@ i.e.~it is really a topological property.
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considered as a topological space.
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\end{itemize}
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\end{example}
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}{}
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\gist{%
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Polish spaces behave very nicely.
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We will see that uncountable polish spaces have size $2^{\aleph_0}$. % TODO: mathfrak c for continuum
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@ -75,11 +76,13 @@ However the converse of this does not hold.
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\item \vocab{Lindelöf} (every open cover has a countable subcover).
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\end{itemize}
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\end{fact}
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\gist{%
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\begin{fact}
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Compact Hausdorff spaces are \vocab{normal} (T4)
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i.e.~two disjoint closed subsets can be separated
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by open sets.
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\end{fact}
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}{}
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\begin{fact}
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For a metric space, the following are equivalent:
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\begin{itemize}
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@ -172,6 +175,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
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D\left( (x_n), (y_n) \right) \coloneqq
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\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
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\]
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\gist{%
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Clearly $D \le 1$.
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It is also clear, that $D$ is a metric.
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@ -180,6 +184,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
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Consider the pointwise limit $(a_n)$.
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This exists since $x_n^{(k)}$ is Cauchy for every fixed $n$.
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Then $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
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}{Clearly $D$ is a complete metric.}
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\end{proof}
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\begin{definition}[Our favourite Polish spaces]
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@ -59,7 +59,6 @@
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define extension, initial segments
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and concatenation of a finite sequence with an infinite one.
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\end{notation}
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}{}
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\begin{definition}
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A \vocab{tree}
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@ -87,6 +86,7 @@
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\forall t\in T.\exists s \supsetneq t.~s \in T.
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\]
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\end{definition}
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}{}
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\begin{definition}
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A \vocab{Cantor scheme}
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@ -139,7 +139,7 @@
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\subsection{The hierarchy of Borel sets}
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Let $\omega_1$ be the first uncountable ordinal.
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\gist{Let $\omega_1$ be the first uncountable ordinal.}{}
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For every $d < \omega_1$,
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we define by transfinite recursion
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classes $\Sigma^0_\alpha$
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@ -201,6 +201,7 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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\gist{%
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\begin{enumerate}[(a)]
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\item \begin{observe}
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\label{ob:sigmasuffices}
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@ -215,7 +216,7 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
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since $\Delta^0_\eta(X)$ is closed under complements.
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Furthermore, it suffices to show $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$,
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by \yaref{ob:sigmasuffices}
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by the observation
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(since $\Sigma^0_\eta(X) \subseteq \Pi^0_\xi(X)$
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and $\Delta^0_\xi(X) = \Sigma^0_\xi(X) \cap \Pi^0_\xi(X)$).
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@ -246,6 +247,17 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
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Then $\bigcup_{n < \omega} A_n \in \Sigma^0_\alpha(X)$.
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It is clear that $\cB_0$ is closed under complements.
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\end{enumerate}
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}{
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\begin{enumerate}[(a)]
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\item It suffices to show that $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$
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for all $1 \le \eta < \xi < \omega_1$.
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For $\eta = 1, \xi = 2$ this holds,
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since open sets of a metrizable space are $F_\sigma$.
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Induction.
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\item Let $\cB_0 \coloneqq \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
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This is a $\sigma$-algebra containing all open sets.
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\end{enumerate}
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}
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\end{proof}
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@ -6,6 +6,7 @@
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% $\fc := 2^{\aleph_0}$
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\end{proposition}
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\begin{proof}
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\gist{%
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We use strong induction on $\xi < \omega_1$.
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We have $\Sigma^0_1(X) \le \fc$
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(for every element of the basis, we can decide
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@ -27,6 +28,9 @@
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\[
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|\cB(X)| \le \omega_1 \cdot \fc = \fc.
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\]
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}{Use strong induction. $|\Sigma^0_1(X)| \le \fc$, since $X$ is second countable.
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\[|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}\]
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}
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\end{proof}
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\begin{proposition}[Closure properties]
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@ -41,6 +41,7 @@ where $X$ is a metrizable, usually second countable space.
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similarly for $\Pi^0_\xi(X)$.
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\end{theorem}
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\begin{proof}
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\gist{%
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Note that if $\cU$ is $2^{\omega}$ universal for
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$\Sigma^0_\xi(X)$, then $(2^{\omega} \times X) \setminus \cU$
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is $2^{\omega}$-universal for $\Pi^0_\xi(X)$.
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@ -89,6 +90,18 @@ where $X$ is a metrizable, usually second countable space.
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Let $A \in \Sigma^0_\xi(X)$.
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Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
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Furthermore $\cU \in \Sigma^0_{\xi}((2^{\omega \times \omega} \times X)$.
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}{
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\begin{itemize}
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\item Suffices for $\Sigma^0_\xi$ (complement is $\Pi^0_\xi$-universal).
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\item $2^\omega$-universal set for $\Sigma^0_1(X)$, since
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$X$ is second countable ($(y,x) \in \cU \iff x \in \bigcup_n \{V_n : y_n = 1\}$).
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\item Induction: Take $\xi_k \to \xi$,
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and $\cU_{\xi_k}$ $\Sigma^0_{\xi_k}$-universal.
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Construct $2^{ \omega \times \omega}$-universal:
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$(y_{m,n}, x) \in \cU :\iff \exists n.~((y_{m,n}), x) \in \cU_{\xi_n}$.
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\end{itemize}
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}
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\end{proof}
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\begin{remark}
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Since $2^{\omega}$ embeds
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@ -96,6 +109,6 @@ where $X$ is a metrizable, usually second countable space.
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% such that the image is closed,
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we can replace $2^{\omega}$ by $Y$
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in the statement of the theorem.%
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\footnote{By definition of the subspace topology
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and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
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\gist{\footnote{By definition of the subspace topology
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and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}}{}
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\end{remark}
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@ -108,14 +108,14 @@
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\end{theorem}
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\begin{proof}
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Let $\phi\colon R \to \Ord$
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by a $\Pi^1_1$-rank.
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Set
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be a $\Pi^1_1$-rank.
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\gist{Set
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\begin{IEEEeqnarray*}{rCl}
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(x,n) \in R^\ast &:\iff& (x,n) \in R\\
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&&\land \forall m.~(x,n) \le^\ast_\phi (x,m)\\
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&&\land \forall m.~\left( (x,n) <^\ast_\phi (x,m) \lor n \le m \right),
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\end{IEEEeqnarray*}
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i.e.~take the element with minimal rank
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i.e.~take}{Take} the element with minimal rank
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that has the minimal second coordinate among those elements.
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\end{proof}
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\gist{
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@ -196,8 +196,8 @@ Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
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\[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
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\begin{itemize}
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\item $n < m$ and
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\item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.%
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\footnote{sic! (there is a typo in the official notes)}
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\item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
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% \footnote{sic! (there was a typo in the official notes)}
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\end{itemize}
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\begin{claim}
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$\prec^\ast$ is well-founded.
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@ -66,7 +66,7 @@
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% \subsection*{Basic Definitions}
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% TODO: move to appendix?
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\gist{%
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Recall:
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\begin{definition}+
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Let $X$ be a set.
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@ -112,6 +112,7 @@ Recall:
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g&\longmapsto & (x \mapsto g \cdot x).
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\end{IEEEeqnarray*}
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\end{remark}
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}{}
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\begin{definition}+
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A group $G$ with a topology
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@ -185,13 +186,13 @@ Recall:
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a homeomorphism $X \leftrightarrow Y$
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commuting with the group action.
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\end{definition}
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\gist{%
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\begin{warning}+
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What is called ``factor'' here is called ``subflow''
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by Furstenberg.
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\end{warning}
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\begin{example}
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Recall that $S_1 = \{z \in \C : |z| = 1\}$.
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Let $X = S_1$, $T = S_1$
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@ -200,6 +201,7 @@ Recall:
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and $\alpha + \beta$ denotes the addition of \emph{angles},
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i.e.~$\alpha \cdot \beta$ in complex numbers.}
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\end{example}
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}{}
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\begin{definition}
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\label{def:isometricextension}
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@ -237,6 +239,7 @@ Recall:
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An isometric extension of a distal flow is distal.
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\end{proposition}
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\begin{proof}
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\gist{%
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Let $\pi\colon X\to Y$ be an isometric extension.
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Towards a contradiction,
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suppose that $x_1,x_2 \in X$ are proximal.
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@ -253,6 +256,11 @@ Recall:
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we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$.
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Therefore $\rho(x_1,x_2) = 0$.
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Hence $x_1 = x_2$ $\lightning$.
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}{Let $\pi\colon X \to Y$ isometric extension.
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Suppose $x_1,x_2 \in X$ is proximal.
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Then $\pi(x_1) = \pi(x_2)$.
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But there exists a $T$-equivariant metric on the fibers.
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}
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\end{proof}
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\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
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bet the quotient space.
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It is compact, second countable and Hausdorff.
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Let $\pi\colon X\to M$ denote the quotient map.
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\gist{%
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\item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$
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is an isometric flow:
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\begin{enumerate}
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@ -204,6 +205,7 @@ i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
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(\yaref{thm:usccomeagercont})
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this implies that $X \setminus \{x_2\}$ is meager.
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But then $X = \{\star\} \lightning$.
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}{}
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\end{enumerate}
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\end{proof}
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@ -173,6 +173,7 @@ More generally we can show:
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then $(Y,T)$ is an isometric extension of $(Z_2, T)$.
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\end{lemma}
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\begin{proof}
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\gist{%
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% TODO TODO TODO Think about this
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For $z_1,z_1' \in Z_1$ with
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$w_1(z_1) = w_1(z_1')$ let
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@ -185,9 +186,16 @@ More generally we can show:
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on the fibers of $Y$ over $Z_2$
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and invariant under $T$.
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$\sigma$ is a metric,
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$\sigma$ is a metric on fibers,
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since if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$,
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then $\pi_1(y) = \pi_1(y')$ or $y = y'$.
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}{%
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\begin{itemize}
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\item Let $\rho\colon Z_1 \times_W Z_1 \to \R$.
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\item Consider $\sigma\colon Y \times_{Z_2} Y \to \R$
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given by $\sigma(y,y') \coloneqq \rho(\pi_1(y), \pi_1(y'))$.
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\end{itemize}
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}
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\end{proof}
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@ -242,7 +250,7 @@ More generally we can show:
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$(X'_\xi, T) = \theta((X_\xi, T)$.
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Let $\pi_\xi$ and $\pi'_\xi$ denote the maps from $X$ to $X_\xi$ resp.~$X'_\xi$.
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Set
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\[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\}.\]
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\[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\} \subseteq X \times X'_{\xi+1}.\]
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Then
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% 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 $n$
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coordinates.
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% TODO ANKI-MARKER
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\begin{lemma}
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\label{lem:lec20:1}
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Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$
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@ -66,7 +63,7 @@ coordinates.
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where $d$ is the metric on $X$,
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$d((x_i), (y_i)) = \max_n \frac{1}{2^n} | x_n - y_n|$.% TODO use multiplicative notation
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\end{lemma}
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\begin{proof}
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\begin{refproof}{lem:lec20:1}
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Let
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\begin{IEEEeqnarray*}{rCl}
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x &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha_{n+1}, \alpha_{n+2},\ldots)\\
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@ -74,7 +71,7 @@ coordinates.
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\end{IEEEeqnarray*}
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We will choose $x_k$ of the form
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\[
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(\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1} \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
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(\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
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\]
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where $\beta_k$ is such that $\frac{\beta_k}{\pi}$ is irrational
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and $|\beta_k| < 2^{-k}$.
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@ -89,7 +86,6 @@ coordinates.
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$u' = (\xi'_n)_{n \in \N}$,
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let $\frac{u}{u'} = (\frac{\xi_n}{\xi'_n})_{n \in \N}$
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($X$ is a group).
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We are interested in $F(x_k, x') = \inf_m d(\tau^m x_k, \tau^m x')$,
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but it is easier to consider the distance between
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their quotient and $1$.
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@ -97,6 +93,7 @@ coordinates.
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\[
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w_k \coloneqq \frac{x_k}{x'} = (\underbrace{1,\ldots,1}_{n-1}, e^{\i \beta_k}, \overbrace{\frac{\alpha_{n+1}}{\alpha'_{n+1}}, \frac{\alpha_{n+2}}{\alpha'_{n+2}}, \ldots}^{\mathclap{\text{not interesting}}}).
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\]
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\gist{%
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\begin{claim}
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$F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
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where $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
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@ -121,7 +118,8 @@ coordinates.
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&\le & 2^{-n} | e^{\i \beta_k}- 1|\\
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&<& 2^{-n-k}.
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\end{IEEEeqnarray*}
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\end{proof}
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}{[some technical details omitted]}
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\end{refproof}
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\begin{definition}
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|
@ -1,6 +1,7 @@
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\lecture{22}{2024-01-16}{}
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\begin{refproof}{thm:21:xnmaxiso}
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% TODO TODO TODO
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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}
|
||||
|
||||
|
||||
|
Loading…
Reference in New Issue
Block a user