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@ -12,6 +12,7 @@
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\end{definition}
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\begin{proposition}
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\label{prop:ifs11}
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$\IF \in \Sigma^1_1(\Tr)$.
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\end{proposition}
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\begin{proof}
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@ -197,16 +197,9 @@ However if we pick $y \in Y$, $Ty$ might not be dense.
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% so $Gy \subseteq Gx$.
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% Since $y = g_0 x \implies x = g_0^{-1}y$, we also have $x \in Gy$,
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% hence $Gx \subseteq Gy$
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% TODO: WHY?
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\end{proof}
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\begin{corollary}
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If $(X,T)$ is distal and minimal,
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then $E(X,T) \acts X$ is transitive.
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\end{corollary}
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@ -125,7 +125,7 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
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% Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
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% with $Y \subseteq G \subseteq \overline{Y}$.
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% $\tilde{f}$ and $\id_G$ agree on $Y$.
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% $Y$ is dense in $G$ and the codomain of $f$ is ltd.\todo{????}
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% $Y$ is dense in $G$ and the codomain of $f$ is Hausdorff.\todo{????}
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% So $f = \id_G$, i.e.~$G = Y$.
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\end{proof}
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@ -109,8 +109,6 @@ for some $B_i \in \cB(Y_i)$.
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\end{itemize}
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\nr 3
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\todo{Wait for mail}
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\begin{lemma}
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Let $X$ be a second-countable topological space.
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Then every base of $X$ contains a countable subset which
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@ -134,6 +132,9 @@ for some $B_i \in \cB(Y_i)$.
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such that $m \in M_{f(m)}$.
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Then $\bigcup_{m \in M} B_{f(m)} = C_n$.
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\end{proof}
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\begin{remark}
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We don't actually need this.
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\end{remark}
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\begin{itemize}
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\item We use the same construction as in exercise 2 (a)
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@ -188,31 +189,71 @@ for some $B_i \in \cB(Y_i)$.
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\]
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is countable as a countable union of countable sets $\lightning$.
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\item Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
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as in the first part.
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Clearly $f$ is also continuous with respect to the new topology,
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so we may assume that $X$ is zero dimensional.
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Let $W \subseteq X$ be such that $f\defon{W}$ is injective
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and $f(W) = f(X)$ (this exists by the axiom of choice).
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Since $f(X)$ is uncountable, so is $W$.
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By the second point, there exist disjoint clopen sets
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$U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
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are uncountable.
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Inductively construct $U_s$ for $s \in 2^{<\omega}$
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as follows:
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Suppose that $U_{s}$ has already been chosen.
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Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
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be disjoint clopen such that $U_{s\concat 1} \cap W$
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and $U_{s\concat 0} \cap W$ are uncountable.
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Such sets exist, since $ U_s \cap W$ is uncountable
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and $U_s$ is a zero dimensional space with the subspace topology.
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And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
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in $U_s$ iff it is clopen in $X$.
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Other proof (without using the existence of a countable clopen
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basis):
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We can cover $X$ by countably many clopen sets of diameter
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$< \frac{1}{n}$:
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Cover $X$ with open balls of diameter $< \frac{1}{n}$.
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Write each open ball as a union of clopen sets.
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That gives us a cover by clopen sets of diameter $< \frac{1}{n}$.
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As $X$ is Lindelöf, there exists a countable subcover.
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Then continue as in the first proof.
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\item Note that this step does not help us to prove the statement.
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It was an error on the exercise sheet.
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% Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
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% as in the first part.
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% Clearly $f$ is also continuous with respect to the new topology,
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% so we may assume that $X$ is zero dimensional.
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%
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% Let $W \subseteq X$ be such that $f\defon{W}$ is injective
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% and $f(W) = f(X)$ (this exists by the axiom of choice).
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% Since $f(X)$ is uncountable, so is $W$.
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% By the second point, there exist disjoint clopen sets
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% $U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
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% are uncountable.
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% Inductively construct $U_s$ for $s \in 2^{<\omega}$
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% as follows:
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% Suppose that $U_{s}$ has already been chosen.
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% Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
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% be disjoint clopen such that $U_{s\concat 1} \cap W$
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% and $U_{s\concat 0} \cap W$ are uncountable.
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% Such sets exist, since $ U_s \cap W$ is uncountable
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% and $U_s$ is a zero dimensional space with the subspace topology.
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% And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
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% in $U_s$ iff it is clopen in $X$.
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Clearly this defines a Cantor scheme.
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\item \todo{TODO}
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\item Let $Y$ be a Polish space and $A \subseteq Y$ analytic
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and uncountable.
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Expand the topology on $Y$ so that $Y$ is zero dimensional
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and $A$ is still analytic.
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Then there exists a Polish space $X$
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and a continuous function $f\colon X \to Y$
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such that $f(X) = A$.
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$A$ is uncountable, so by (2)
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there exists non-empty disjoint clopen $V_0$, $V_1$
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such that $V_0 \cap A$ and $V_1 \cap A$
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are uncountable.
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Let $W_0 = f^{-1}(V_0 \cap A)$ and $W_1 = f^{-1}(V_1 \cap A)$.
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$ W_0$ and $W_1$ are clopen and disjoint.
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We can cover $W_0$ with countably many open sets
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of diameter $\le \frac{1}{n}$
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and similarly for $W_1$.
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Then pick open sets such that there image is uncountable.
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Repeating this construction we get a Cantor scheme on $X$.
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So we get $2^{\N} \overset{s}{\hookrightarrow} X$
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and by construction of the cantor scheme,
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we get that $f \circ s$ is injective and continuous.
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\end{itemize}
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\nr 4
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159
inputs/tutorial_09.tex
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159
inputs/tutorial_09.tex
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@ -0,0 +1,159 @@
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\tutorial{09}{2023-12-12}{}
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\begin{fact}
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Let $X,Y$ be topological spaces
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$X$ (quasi-)compact and
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$Y$ Hausdorff.
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Let $f\colon X\to Y$ be a continuous bijection.
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Then $f$ is a homeomorphism.
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\end{fact}
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\begin{proof}
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Compact subsets of Hausdorff spaces are closed.
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\end{proof}
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\subsection{Sheet 8}
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Material on topological dynamic
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\begin{itemize}
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\item Terence Tao's notes on ergodic theory 254 A.
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\item Furstenberg (uses very different notation!).
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\end{itemize}
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\nr 1
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\begin{remark}
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$\Sigma^1_1$-complete sets are in some sense the ``worst''
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$\Sigma^1_1$-sets:
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Deciding whether an element is contained in the
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$\Sigma^1_1$-complete set
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is at least as ``hard'' as as for any $\Sigma^1_1$ set.
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In particular, $\Sigma^1_1$-complete sets
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are not Borel.
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\end{remark}
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Similarly as in \yaref{prop:ifs11} it can be shown that $L \in \Sigma^1_1(X)$:
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Consider $\{(x, \beta) \in X \times \cN : \forall n.~x_{\beta_n} | x_{\beta_{n+1}}\}$.
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This is closed in $X \times \cN$, since it is a countable intersection of clopen
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sets and $L = \proj_X(D)$.
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Since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete,
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it suffices to find a Borel map $f\colon \Tr \to X$
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such that $x \in \IF \iff f(x) \in L$.
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Let $\phi\colon \omega^2 + \omega \to \omega$ be bijective
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and let $p_i$ denote the $i$-th prime.
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Define
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\begin{IEEEeqnarray*}{rCl}
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\psi\colon \omega^{<\omega} &\longrightarrow & \omega \setminus \{0\} \\
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(s_0, s_1, \ldots, s_{n-1})&\longmapsto & \prod_{i < n} p_{\phi(\omega \cdot i + s_i)}.
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\end{IEEEeqnarray*}
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Note that $\psi$ is injective and that $s \in \omega^{<\omega}$ is an initial segment
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of $t \in \omega^{<\omega}$ iff $\psi(s) | \psi(t)$.
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Let
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\begin{IEEEeqnarray*}{rCl}
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f' \colon \Tr &\longrightarrow & \cP(\omega \setminus \{0\}) \\
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T &\longmapsto & \{\phi(s) : s \in T\}.
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\end{IEEEeqnarray*}
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We can turn this into a function
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$f\colon \Tr \to (\omega \setminus \{0\})^{\omega}$
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by mapping a subset of $\omega \setminus \{0\}$
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to the unique strictly increasing sequence whose elements are from
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that subset
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(appending $\phi(\omega^2 + n), n \in \omega$, if the subset was finite).
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Note that $T \in \IF \iff f(T) \in L$.
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Furthermore $f$ is Borel, since fixing
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a finite initial sequence (i.e.~a basic open set of $(\omega \setminus \{0\})^{\omega}$)
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amounts to a finite number of conditions on the preimage.
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\nr 2
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\todo{handwritten}
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% Aron
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%
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% \begin{enumerate}[1.]
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% \item This is trivial $\sup \sup$.
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% \item Clearly there are trees of rank $n$ for all $n < \omega$.
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% Glue them together.
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% \[
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% \{(0,0,i,\underbrace{0,\ldots,0}_{i \text{~times}}) | i < \omega\}.
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% \]
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% \item Map infinite branches. $\sup$ the $\le $.
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% \item Induction on $\rho(S)$.
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% Cofinal subsequences bla bla.
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% \end{enumerate}
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\nr 3
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\begin{itemize}
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\item $LO(\N) \overset{\text{closed}}{\subseteq} 2^{\N\times \N}$:
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We have $< \in LO(\N)$ iff
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\begin{itemize}
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\item $\forall x,y.~ (x \neq y \implies (x< y \lor x > y))$,
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\item $\forall x.~(x \not < x)$,
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\item $\forall x,y,x.~(x < y < z \implies x < z)$.
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\end{itemize}
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Write this with $\bigcap$,
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i.e.
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\begin{IEEEeqnarray*}{rCl}
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LO(\N) &=& \bigcap_{n \in \N} \{R: (n,n) \not\in R\}\\
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&& \cap \bigcap_{m < n \in \N} (\{R: (n,m) \in R\} \cup \{R: (m,n) \in R\})\\
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&&\cap \bigcap_{a,b,c \in \N} (\{R: (a,b) \in R \land (b,c) \in R \implies (a,c) \in R\}.
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\end{IEEEeqnarray*}
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This is closed as an intersection of clopen sets.
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\item We apply \yaref{thm:borel} (iv).
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Let $\cF \subseteq LO(\N) \times \cN$ be
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such that the $\cN$-coordinate encodes a strictly
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decreasing sequence, i.e.~
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\[(R, s) \in \cF :\iff \forall n \in \N.~(s(n+1), s(n)) \in R.\]
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We have that
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\[
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\cF = \bigcap_{n \in \N} \{(R,s) \in LO(\N)\times \cN : (s(n+1), s(n)) \in R\}
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\]
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is closed as an intersection of clopen sets.
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Clearly $\pr_{LO(\N)}(\cF)$ is the complement
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of $WO(\N)$, hence $WO(\N)$ is coanalytic.
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\end{itemize}
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\nr 4
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\begin{remark}
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In the lecture we only look at metrizable flows,
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so the definitions from the exercise sheet and from
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the lecture don't agree.
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\end{remark}
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\begin{itemize}
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\item Consider
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\begin{IEEEeqnarray*}{rCl}
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\Phi\colon \Z\text{-flows on } X &\longrightarrow & \Homeo(X) \\
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(\alpha\colon \Z\times X \to X) &\longmapsto & \alpha(1, \cdot)\\
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\begin{pmatrix*}[l]
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\Z\times X &\longrightarrow & X \\
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(z,x) &\longmapsto & \beta^{z}(x)
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\end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X).
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\end{IEEEeqnarray*}
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Clearly this has the desired properties.
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\item We have
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\begin{IEEEeqnarray*}{Cl}
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& \Z \circlearrowright X \text{ has a dense orbit}\\
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\iff& \exists x \in X.~ \overline{\Z\cdot x} = X\\
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\iff& \exists x \in X.~\forall U\overset{\text{open}}{\subseteq} X.~\exists z \in \Z.~
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z \cdot x \in U\\
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\iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~
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\exists z \in \Z.~f^z(x) \in U.
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\end{IEEEeqnarray*}
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\item Not solved.
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\item It suffices to check the condition from part (b)
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for open sets $U$ of a countable basis
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and points $x \in X$ belonging to a countable dense subset.
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Replacing quantifiers by unions resp.~intersections
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gives that $D$ is Borel.
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\end{itemize}
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@ -134,6 +134,9 @@
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\DeclareSimpleMathOperator{WO} % well orderings
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\DeclareSimpleMathOperator{Homeo}
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\DeclareSimpleMathOperator{osc} % oscillation
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\newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}
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\input{inputs/tutorial_06}
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\input{inputs/tutorial_07}
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\input{inputs/tutorial_08}
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\input{inputs/tutorial_09}
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\section{Facts}
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\input{inputs/facts}
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