Josia Pietsch 24ed36d0a7
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\subsection{Sheet 6}
% 5 / 20
\nr 1
Note that not every set has a density!
\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,
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.
\nr 2
Recall \yaref{thm:clopenize}:
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).
\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
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 $\tau_Y^\omega$.
If you do something and it didn't work,
try doing it again ($\omega$-many times).
\nr 3
\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
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).
\nr 4
\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.
\begin{theorem}\footnote{See Kechris 16.1}
Let $X, Y$ be Polish.
For $\emptyset \neq U \overset{\text{open}}{\subseteq} Y$
\[A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\}.\]
\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ A_U \text{ is Borel}\}.\]
Then $\cA$ contains all Borel sets.
\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:
&& \{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
(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 $U \overset{\text{open}}{\subseteq} V$).