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178 lines
6.1 KiB
TeX
178 lines
6.1 KiB
TeX
\subsection{Sheet 6}
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\tutorial{07}{2023-11-28}{}
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% 5 / 20
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\nr 1
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\begin{warning}
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Note that not every set has a density!
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\end{warning}
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\begin{enumerate}[(a)]
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\item Let $X = \bI^{\omega}$.
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Let $C_0 = \{(x_n) : x_n \to 0\}$.
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Claim: $C_0 \in \Pi^0_3(X)$ (intersections of $F_\sigma$ sets).
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We have
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\[
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x \in C_0 \iff \forall q \in \Q^+.~\exists N.~\forall n \ge N.~x_n \le q,
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\]
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i.e.
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\[
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C_0 = \bigcap_{q \in \Q^+}\bigcup_{N < \omega} \bigcap_{n > N} \{x_n : x_n \le q\}.
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\]
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Clearly this is a $\Pi^0_3$ set.
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\item Let $Z \coloneqq \{f \in 2^{\omega} : f(\N) \text{ has density 0}\}$.
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Claim: $Z \in \Pi^0_3(2^{\N})$.
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It is
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\[
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Z = \bigcap_{q \in \Q^+} \bigcup_{N < \omega}
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\bigcap_{n \ge N}\{f \in 2^{\omega} : \frac{\sum_{i < n} f(i)}{n} \le nq\}.
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\]
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Clearly this is a $\Pi^0_3$-set.
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\end{enumerate}
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\nr 2
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Recall \yaref{thm:clopenize}:
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\begin{fact}
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Let $(X,\tau)$ be a Polish space and
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$A \in \cB(X)$.
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Then there exists $\tau' \supseteq \tau$
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with the same Borel sets as $\tau$
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such that $A$ is clopen.
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\end{fact}
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(Do it for $A$ closed,
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then show that the sets which work
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form a $\sigma$-algebra).
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\begin{enumerate}[(a)]
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\item Let $(X, \tau)$ be Polish.
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We want to expand $\tau$ to a Polish topology
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$\tau_0$ maintaining the Borel sets,
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such that $(X, \tau')$ is 0d.
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Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$.
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Each $U_n$ is open, hence Borel,
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so by \hyperref[thm:clopenize]{a theorem from the lecture™}
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there exists a Polish topology $\tau_n$
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such that $U_n$ is clopen, preserving Borel sets.
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Hence we get $\tau_\infty$
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such that all the $V_n$ are clopen in $\tau_\infty$.
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Let $\tau^{1} \coloneqq \tau_\infty$.
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Do this $\omega$-many times to get $\tau^{\omega}$.
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$\tau^{\omega}$ has a base consisting
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of finite intersections $A_1 \cap \ldots \cap A_n$,
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where $A_i$ is a basis element we chose
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to construct $\tau_i$,
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hence clopen.
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\item Let $(X, \tau_X), Y$ be Polish
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and $f\colon X \to Y$ Borel.
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Show $\exists \tau' \supseteq \tau$ maintaining the Borel structure
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with $f$ continuous.
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Let $(U_n)_n$ be a countable base of $Y$.
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Clopenize all the preimages of the $(U_n)_n$.
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\item Let $f\colon X \to Y$ be a Borel isomorphism.
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Then there are finer topologies preserving the Borel
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structure
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such that $f\colon X' \to Y'$ is a homeomorphism.
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Repeatedly apply (c).
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Get $\tau_X^1$ to make $f$ continuous.
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Then get $\tau_Y^1$ to make $f^{-1}$ continuous
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(possibly violating continuity of $f$)
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and so on.
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Let $\tau_X^\omega \coloneqq \langle \tau_X^n \rangle$
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and similarly for $\tau_Y^\omega$.
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\end{enumerate}
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\begin{idea}
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If you do something and it didn't work,
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try doing it again ($\omega$-many times).
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\end{idea}
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\nr 3
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\begin{enumerate}[(a)]
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\item Show that if $\Gamma$ is self-dual (closed under complements)
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and closed under continuous preimages,
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then for any topological space $X$,
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there does not exist an $X$-universal set for $\Gamma(X)$.
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Suppose there is an $X$-universal set for $\Gamma(X)$,
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i.e.~$U \subseteq X \times X$
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such that $U \in \Gamma(X \times X) \land \{U_x : \in X\} = \Gamma(X)$.
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Consider $X \xrightarrow[x\mapsto (x,x)]{d} X \times X$.
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Let $V = U^c$.
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Then $V \in \Gamma(X \times X)$ and $d^{-1}(V) \in \Gamma(X)$.
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Then $d^{-1}(V) = U_x$ for some $x$.
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But then $(x,x) \in U \iff x \in d^{-1}(V) \iff (x,x) \not\in U \lightning$.
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\item Let $\xi$ be an ordinal
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and let $X$ be a topological space.
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Show that neither $\cB(X)$ nor $\Delta^0_\xi(X)$ can have $X$-universal
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sets.
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Clearly $\cB(X)$ is self-dual and closed under continuous preimages.
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Clearly $\Delta^0_\xi(X)$ is self-dual
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and closed under continuous preimages (by a trivial induction).
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\end{enumerate}
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\nr 4
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Recall:
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\begin{fact}[Sheet 5, Exercise 1]
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Let $\emptyset\neq X$ be a Baire space.
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Then $\forall A \subseteq X$,
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$A$ is either meager or locally comeager.
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\end{fact}
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\begin{theorem}\footnote{See Kechris 16.1}
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Let $X, Y$ be Polish.
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For $\emptyset \neq U \overset{\text{open}}{\subseteq} Y$
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let
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\[A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\}.\]
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Define
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\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ A_U \text{ is Borel}\}.\]
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Then $\cA$ contains all Borel sets.
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\end{theorem}
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\begin{proof}
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\begin{enumerate}[(i)]
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\item Show for $V \in \cB(X), W \overset{\text{open}}{\subseteq} Y$
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that $V \times W \in \cA$.
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Clearly $V \times W$ is Borel
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and $\{x \in X: W \cap U \text{ is not meager}\} \in \{\emptyset, V\}$.
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\item Let $(A_n)_{n < \omega} \in \cA^{\omega}$.
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Then $\bigcap_n A_n \in \cA$.
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($(\bigcup_n A_n)_U = \bigcup_n (A_{n})_U$).
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\item Let $A \in \cA$ and $B = A^c$.
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Fix $\emptyset\neq U \subseteq Y$.
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Then $\{x : A_x \text{is not meager in $U$}\}$ is Borel,
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i.e.~$\{x : A_x^c \text{ is not meager in $U$}\}$ is Borel.
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Since $A$ is Borel, $A_x$ is Borel as well.
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Hence by the fact:
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\begin{IEEEeqnarray*}{rCl}
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&& \{x : A_x^c \text{ is not meager in $U$}\}\\
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&=& \{x \colon A_x^c \text{ is locally comeager in $U$}\}\\
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&=& \{x \colon \exists \emptyset\neq V \overset{\text{open}}{\subseteq} V.~ A_x \text{ is meager in $V$}\}\\
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&=& \bigcup_{\emptyset \neq V \overset{\text{open}}{\subseteq} U} A_V^c
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\end{IEEEeqnarray*}
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(a countable union suffices, since we only need to check this for $V$ of the basis; if $A \subseteq V$ is nwd, then $A \cap U \subseteq U$ is nwd for all $U \overset{\text{open}}{\subseteq} V$).
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\end{enumerate}
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\end{proof}
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