Josia Pietsch
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\subsection{Sheet 6}


\tutorial{07}{20231128}{}




% 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 selfdual (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 selfdual and closed under continuous preimages.


Clearly $\Delta^0_\xi(X)$ is selfdual


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}
