w23-logic-3/inputs/tutorial_07.tex

\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
\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}

\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}