diff --git a/inputs/tutorial_06.tex b/inputs/tutorial_06.tex new file mode 100644 index 0000000..577ef2c --- /dev/null +++ b/inputs/tutorial_06.tex @@ -0,0 +1,180 @@ +\tutorial{06}{2023-11-28}{} + +% 5 / 20 + +\subsection{Exercise 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} + +\subsection{Exercise 2} + +\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. + + (Do it for $A$ closed, + then show that the sets which work + form a $\sigma$-algebra). +\end{fact} + +\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 a theorem from the lecture$^{\text{tm}}$ + 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} + +\subsection{Exercise 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} + +\subsection{Exercise 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}[Kechris 16.1] + Let $X, Y$ be Polish. + + Let + \[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ + A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\] + + 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} + + + + + +