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