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1cc501fbe3
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a4ac283146
9 changed files with 14 additions and 126 deletions
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@ -197,7 +197,8 @@
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Let $X \neq \emptyset$ be a Polish space.
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Let $X \neq \emptyset$ be a Polish space.
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Then there is a closed subset
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Then there is a closed subset
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\[
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\[
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D \subseteq \N^\N \mathbin{\text{\reflectbox{$\coloneqq$}}} \cN
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D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN
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% TODO correct N for the Baire space?
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\]
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\]
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and a continuous bijection from
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and a continuous bijection from
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$D$ onto $X$ (the inverse does not need to be continuous).
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$D$ onto $X$ (the inverse does not need to be continuous).
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@ -238,7 +239,7 @@
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\end{enumerate}
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\end{enumerate}
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\gist{%
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\gist{%
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Suppose we already have $F_s \mathbin{\text{\reflectbox{$\coloneqq$}}} F$.
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Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
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We need to construct a partition $(F_i)_{i \in \N}$
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We need to construct a partition $(F_i)_{i \in \N}$
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of $F$ with $\overline{F_i} \subseteq F$
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of $F$ with $\overline{F_i} \subseteq F$
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and $\diam(F_i) < \epsilon$
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and $\diam(F_i) < \epsilon$
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@ -62,7 +62,7 @@
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We extend $f$ to $g\colon\cN \to X$
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We extend $f$ to $g\colon\cN \to X$
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in the following way:
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in the following way:
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Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x\defon{n} = s\}$.
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Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
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Clearly $S$ is a pruned tree.
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Clearly $S$ is a pruned tree.
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Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1})
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Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1})
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\[
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\[
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@ -4,7 +4,6 @@
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If $C$ is coanalytic,
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If $C$ is coanalytic,
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then there exists a $\Pi^1_1$-rank on $C$.
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then there exists a $\Pi^1_1$-rank on $C$.
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\end{theorem}
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\end{theorem}
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% TODO show that WO sse 2^QQ is Pi_1^1 complete
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\begin{proof}
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\begin{proof}
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\gist{%
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\gist{%
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Pick a $\Pi^1_1$-complete set.
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Pick a $\Pi^1_1$-complete set.
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@ -26,7 +25,7 @@
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% \arrow["\subseteq"', hook, from=2-3, to=1-3]
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% \arrow["\subseteq"', hook, from=2-3, to=1-3]
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% \end{tikzcd}
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% \end{tikzcd}
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Let $X = 2^{\Q} \supseteq \WO$.
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Let $X = 2^{\Q} \supseteq \WO$.
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We have already shown that $\WO$ is $\Pi^1_1$-complete.% TODO REF
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We have already show that $\WO$ is $\Pi^1_1$-complete.
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Set $\phi(x) \coloneqq \otp(x)$
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Set $\phi(x) \coloneqq \otp(x)$
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($\otp\colon \WO \to \Ord$ denotes the order type).
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($\otp\colon \WO \to \Ord$ denotes the order type).
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@ -175,7 +175,7 @@ Recall:
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Let $(T,X)$ and $(T,Y)$ be flows.
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Let $(T,X)$ and $(T,Y)$ be flows.
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A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$
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A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$
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is a continuous surjection $X \twoheadrightarrow Y$
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is a continuous surjection $X \twoheadrightarrow Y$
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that is $T$-equivariant,
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commuting with the group action,
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i.e.~$\forall t \in T, x \in X.~\pi(t\cdot x) = t\cdot \pi(x)$.
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i.e.~$\forall t \in T, x \in X.~\pi(t\cdot x) = t\cdot \pi(x)$.
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If such a factor map exists,
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If such a factor map exists,
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we also say that $(T,Y)$ is a \vocab{factor}
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we also say that $(T,Y)$ is a \vocab{factor}
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@ -258,14 +258,13 @@ Recall:
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\begin{definition}
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\begin{definition}
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Let $\Sigma = \{(X_i, T) : i \in I\} $
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Let $\Sigma = \{(X_i, T) : i \in I\} $
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be a collection of factors of $(X,T)$.
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be a collection of factors of $(X,T)$.
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Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
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Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor maps.
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Then $(X, T)$ is a \vocab{limit}%
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Then $(X, T)$ is the \vocab{limit} of $\Sigma$
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\footnote{This is not a limit in the category theory sense and not uniquely determined.}
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of $\Sigma$
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iff
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iff
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\[
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\[
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\forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
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\forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
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\]
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\]
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% TODO think about abstract nonsense
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\end{definition}
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\end{definition}
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\begin{proposition}
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\begin{proposition}
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@ -65,9 +65,9 @@ equicontinuity coincide.
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\end{tikzcd}\]
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\end{tikzcd}\]
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\end{theorem}
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\end{theorem}
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\begin{proof}
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\begin{proof}
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% TODO Think about this
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We want to apply Zorn's lemma.
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We want to apply Zorn's lemma.
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If suffices to show that isometric flows are closed under inverse limits,%
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If suffices to show that isometric flows are closed under inverse limits,
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\footnote{This seems to be an inverse limit in the category theory sense.}
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i.e.~if $(Y_\alpha, f_{\alpha,\beta})$,
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i.e.~if $(Y_\alpha, f_{\alpha,\beta})$,
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$\beta < \alpha \le \Theta$
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$\beta < \alpha \le \Theta$
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are isometric, then the inverse limit $Y$ is isometric.%
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are isometric, then the inverse limit $Y$ is isometric.%
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@ -34,9 +34,6 @@
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\end{enumerate}
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\end{enumerate}
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\nr 2
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\nr 2
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Recall \yaref{thm:clopenize}:
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\begin{fact}
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\begin{fact}
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Let $(X,\tau)$ be a Polish space and
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Let $(X,\tau)$ be a Polish space and
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$A \in \cB(X)$.
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$A \in \cB(X)$.
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@ -44,10 +41,10 @@ Recall \yaref{thm:clopenize}:
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with the same Borel sets as $\tau$
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with the same Borel sets as $\tau$
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such that $A$ is clopen.
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such that $A$ is clopen.
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\end{fact}
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(Do it for $A$ closed,
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(Do it for $A$ closed,
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then show that the sets which work
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then show that the sets which work
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form a $\sigma$-algebra).
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form a $\sigma$-algebra).
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\end{fact}
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\begin{enumerate}[(a)]
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\begin{enumerate}[(a)]
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\item Let $(X, \tau)$ be Polish.
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\item Let $(X, \tau)$ be Polish.
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@ -11,26 +11,6 @@
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\nr 3
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\nr 3
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% somewhat examinable (for 1.0)
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% somewhat examinable (for 1.0)
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% TODO
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\begin{enumerate}[(a)]
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\item $(X,T)$ is distal iff it does not have a proximal pair,
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i.e.~$a\neq b$, $c$ such that $t_n \in T$,
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$t_na, t_nb \to c$.
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Equivalently,
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for all $a,b$ there exists an $\epsilon$,
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such that for all $t \in T$, $d(ta,tb) > \epsilon$.
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\item % TODO (not too hard)
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% (b)
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% Let $(X,T)$ be distal with a dense orbit,
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% then it is distal minimal.
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% Sheet 8: has dense orbit is Borel
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% Distal flow decomposes into distal minimal flows.
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\end{enumerate}
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\nr 4
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\nr 4
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@ -1,87 +0,0 @@
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\tutorial{15}{2024-01-31}{Additions}
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\subsection{Additional Tutorial}
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The following is not relevant for the exam,
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but gives a more general picture.
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Let $ X$ be a topological space.
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Let $\cF$ be a filter on $ X$.
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$x \in X$ is a limit point of $\cF$ iff the neighbourhood filter $\cN_x$,
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all sets containing an open neighbourhood of $x$,
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is contained in $\cF$.
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\begin{fact}
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$X$ is Hausdorff iff every filter has at most one limit point.
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\end{fact}
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\begin{proof}
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Neighbourhood filters are compatible
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iff the corresponding points
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can not be separated by open subsets.
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\end{proof}
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\begin{fact}
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$X$ is (quasi-) compact
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iff every ultrafilter converges.
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\end{fact}
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\begin{proof}
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Suppose that $X$ is compact.
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Let $\cU$ be an ultrafilter.
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Consider the family $\cV = \{\overline{A} : A \in \cU\}$
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of closed sets.
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By the FIP we geht that there exist
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$c \in X$ such that $c \in \overline{A}$ for all $A \in \cU$.
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Let $N$ be an open neighbourhood of $c$.
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If $N^c \in \cU$, then $c \in N^c \lightning$
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So we get that $N \in \cU$.
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Let $\{V_i : i \in I\} $ be a family of closed sets with the FIP.
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Consider the filter generated by this family.
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We extend this to an ultrafilter.
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The limit of this ultrafilter is contained in all the $V_i$.
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\end{proof}
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Let $X,Y$ be topological spaces,
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$\cB$ a filter base on $X$,
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$\cF$ the filter generated by $\cB$
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and
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$f\colon X \to Y$.
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Then $f(\cB)$ is a filter base on $Y$,
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since $f(\bigcap A_i ) \subseteq \bigcap f(A_i)$.
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We say that $\lim_\cF f = y$,
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if $f(\cF) \to y$.
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Equivalently $f^{-1}(N) \in \cF$
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for all neighbourhoods $N$ of $y$.
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In the lecture we only considered $X = \N$.
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If $\cB$ is the base of an ultrafilter,
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so is $f(\cB)$.
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\begin{fact}
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Let $X$ be a topological space
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and let $Y$ be Hausdorff.
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Let $f,g \colon X \to Y$
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be continuous.
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Let $A \subseteq X$ be dense such that
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$f\defon{A} = g\defon{A} $.
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Then $f = g$.
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\end{fact}
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\begin{proof}
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Consider $(f,g)^{-1}(\Delta) \supseteq A$.
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\end{proof}
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We can uniquely extend $f\colon X \to Y$ continuous
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to a continuous $\overline{f}\colon \beta X \to Y$
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by setting $\overline{f}(\cU) \coloneqq \lim_\cU f$.
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Let $V$ be an open neighbourhood of $Y$ in $\overline{f}\left( U) \right) $.
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Consider $f^{-1}(V)$.
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Consider the basic open set
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\[
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\{\cF \in \beta\N : \cF \ni f^{-1}(V)\}.
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\]
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@ -74,7 +74,6 @@
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\input{inputs/tutorial_12b}
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\input{inputs/tutorial_12b}
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\input{inputs/tutorial_12}
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\input{inputs/tutorial_12}
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\input{inputs/tutorial_14}
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\input{inputs/tutorial_14}
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\input{inputs/tutorial_15}
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\section{Facts}
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\section{Facts}
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\input{inputs/facts}
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\input{inputs/facts}
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