w23-logic-3/inputs/lecture_07.tex

\lecture{07}{2023-11-07}{}

\begin{proposition}
    Let $X$ be second countable.
    Then $|\cB(X)| \le \fc$.
    % $\fc := 2^{\aleph_0}$
\end{proposition}
\begin{proof}
    We use strong induction on  $\xi < \omega_1$.
    We have $\Sigma^0_1(X) \le \fc$
    (for every element of the basis, we can decide
    whether to use it in the union or not).

    Suppose that $\forall \xi' < \xi.~|\Sigma^0_{\xi'}(X)| \le \fc$.
    Then $|\Pi^0_{\xi'}(X)| \le  \fc$.
    We have that
    \[
        \Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
    \]
    Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.

    We have
    \[
    \cB(X) = \bigcup_{\xi < \omega_1}  \Sigma^0_\xi(X).
    \]
    Hence
    \[
    |\cB(X)| \le  \omega_1 \cdot  \fc = \fc.
    \]
\end{proof}

\begin{proposition}[Closure properties]
    Suppose that $X$ is metrizable.
    Let $1 \le  \xi < \omega_1$.
    Then
    \begin{enumerate}[(a)]
        \item \begin{itemize}
                \item $\Sigma^0_\xi(X)$ is closed under countable unions.
                \item  $\Pi^0_\xi(X)$ is closed under countable intersections.
                \item  $\Delta^0_\xi(X)$ is closed under complements.
            \end{itemize}
        \item \begin{itemize}
                \item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
                \item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
                \item $\Delta^0_\xi(X)$ is closed under finite unions and
                    finite intersections.
            \end{itemize}

    \end{enumerate}
\end{proposition}
\gist{%
\begin{proof}
    \begin{enumerate}[(a)]
        \item This follows directly from the definition.
            Note that a countable intersection can be written
            as a complement of the countable union of complements:
            \[
            \bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
            \]
        \item If suffices to check this for $\Sigma^0_{\xi}(X)$.
            Let $A = \bigcup_{n}  A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
            and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
            Then
            \[
            A \cap B = \bigcup_{n,m}  \left( A_n \cap B_m \right)
            \]
            and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.

    \end{enumerate}
\end{proof}
}{}
\begin{example}
    Consider the cantor space $2^{\omega}$.
    We have that $\Delta^0_1(2^{\omega})$
    is not closed under countable unions%
    \gist{ (countable unions yield all open sets, but there are open
    sets that are not clopen)}{}.
\end{example}

\subsection{Turning Borels Sets into Clopens}

\begin{theorem}%
    \gist{%
    \footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
            unfortunately seems to be non-standard vocabulary.
            Our tutor repeatedly advised against using it in the final exam.
            Contrary to popular belief
            the very same tutor was \textit{not} the one first to introduce it,
            as it would certainly be spelled ``to clopenise'' if that were the case.
        }%
    }{}%
    \label{thm:clopenize}
    Let $(X, \cT)$ be a Polish space.
    For any  Borel set $A \subseteq  X$,
    there is a finer Polish topology,%
    \footnote{i.e.~$\cT_A \supseteq \cT$ and $(X, \cT_A)$ is Polish}
    such that
    \begin{itemize}
        \item $A$ is clopen in $\cT_A$,
        \item the Borel sets do not change,
            i.e.~$\cB(X, \cT) = \cB(X, \cT_A)$.
    \end{itemize}
\end{theorem}
\begin{corollary}[Perfect set property]
    Let $(X, \cT)$ be Polish,
    and let $B \subseteq X$ be Borel and uncountable.
    Then there is an embedding
    of the cantor space $2^{\omega}$
    into $B$.
\end{corollary}
\begin{proof}
    Pick $\cT_B \supset \cT$
    such that $(X, \cT_B)$ is Polish,
    $B$ is clopen in $\cT_B$ and
    $\cB(X,\cT) = \cB(X, \cT_B)$.

    Therefore $(\cB, \cT_B\defon{B})$ is Polish.
    We know that there is an embedding
    $f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$.

    Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$.
    This is still continuous as $\cT \subseteq \cT_B$.
    Since $2^{\omega}$ is compact, $f$ is an embedding.
    %\todo{Think about this}
\end{proof}

\begin{refproof}{thm:clopenize}
    We show that
    \begin{IEEEeqnarray*}{rCl}
        A \coloneqq  \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\
                                 && (X, \cT_B)  \text{ is Polish},\\
                                 && \cB(X, \cT) = \cB(X, \cT_B)\\
                                 && B \text{ is clopen in $\cT_B$}\\
        \}
    \end{IEEEeqnarray*}
    is equal to the set of Borel sets.

    The proof rests on two lemmata:
    \begin{lemma}
        \label{thm:clopenize:l1}
        Let $(X,\cT)$ be a Polish space.
        Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$)
        there is $\cT_F \supseteq \cT$
        such that $\cT_F$ is Polish,
        $\cB(\cT) = \cB(\cT_F)$
        and  $F$ is clopen in $\cT_F$.
    \end{lemma}
    \begin{proof}
        Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
        Both are Polish spaces.
        Take the coproduct%
        \footnote{In the lecture, this was called the \vocab{topological sum}.}
        $F \oplus (X \setminus F)$ of these spaces.
        This space is Polish,
        and the topology is generated by $\cT \cup \{F\}$,
        hence we do not get any new Borel sets.
    \end{proof}
    So all closed sets are in $A$.
    Furthermore $A$ is closed under complements,
    since complements of clopen sets are clopen.

    \begin{lemma}
        \label{thm:clopenize:l2}
        Let $(X, \cT)$ be Polish.
        Let $\{\cT_n\}_{n < \omega}$
        be Polish topologies
        such that $\cT_n \supseteq \cT$
        and $\cB(\cT_n) = \cB(\cT)$.
        Then the topology $\cT_\infty$ generated by  $\bigcup_{n} \cT_n$
        is Polish
        and $\cB(\cT_\infty) = \cB(T)$.
    \end{lemma}
    \begin{proof}
        We have that $\cT_\infty$ is the smallest
        topology containing all $\cT_n$.
        To get $\cT_\infty$
        consider
        \[
        \cF \coloneqq  \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}.
        \]
        Then
        \[
        \cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}.
        \]
        (It suffices to take countable unions,
        since we may assume that the $A_1, \ldots, A_n$ in the
        definition of $\cF$ belong to
        a countable basis of the respective $\cT_n$).

        % Proof was finished in lecture 8
        Let $Y = \prod_{n \in \N} (X, \cT_n)$.
        Then $Y$ is Polish.
        Let $\delta\colon (X, \cT_\infty) \to Y$
        defined by $\delta(x) = (x,x,x,\ldots)$.
        \begin{claim}
            $\delta$ is a homeomorphism.
        \end{claim}
        \begin{subproof}
            Clearly $\delta$ is a bijection.
            We need to show that it is continuous and open.

            Let $U \in \cT_i$.
            Then
            \[
            \delta^{-1}(D \cap \left( X \times  X \times \ldots\times  U \times  \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
            \]
            hence $\delta$ is continuous.
            Let $U \in \cT_\infty$.
            Then $U$ is the union of sets of the form
            \[
            V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
            \]
            for some $n_1 < n_2 < \ldots < n_u$
            and $U_{n_i} \in \cT_i$.

            Thus is suffices to consider sets of this form.
            We have that
            \[
            \delta(V) = D \cap (X \times  X \times  \ldots \times  U_{n_1} \times  \ldots \times  U_{n_2} \times  \ldots \times  U_{n_u} \times  X \times  \ldots) \overset{\text{open}}{\subseteq} D.
            \]
        \end{subproof}

        This will finish the proof since
        \[
        D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
        \]
        Why? Let  $(x_n) \in Y \setminus D$.
        Then there are $i < j$ such that $x_i \neq x_j$.
        Take disjoint open $x_i \in U$, $x_j \in V$.
        Then
        \[(x_n) \in  X \times  X \times  \ldots \times  U \times  \ldots \times  X \times  \ldots \times V \times  X \times  \ldots\]
        is open in $Y\setminus D$.
        Hence $Y \setminus D$ is open, thus $D$ is closed.
        It follows that $D$ is Polish.
    \end{proof}

    We need to show that $A$ is closed under countable unions.
    By \yaref{thm:clopenize:l2} there exists a topology
    $\cT_\infty$ such that $A = \bigcup_{n < \omega}  A_n$ is open in $\cT_\infty$
    and $\cB(\cT_\infty) = \cB(\cT)$.
    Applying \yaref{thm:clopenize:l1}
    yields a topology $\cT_\infty'$ such that
    $(X, \cT_\infty')$ is Polish,
    $\cB(\cT_\infty') = \cB(\cT)$
    and $A $ is clopen in $\cT_{\infty}'$.
\end{refproof}