\subsection{Sheet 3}

\tutorial{04}{}{}

\nr 1

Let $A \neq \emptyset$ be discrete.
For $D \subseteq  A^{\omega}$,
let
\[
    T_D \coloneqq  \{x\defon{n} \in  A^{<\omega} | x \in D, n \in \N\}.
\]
\begin{enumerate}[(a)]
    \item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:

        Clearly $T_D$ is a tree.
        Let $x \in T_D$.
        Then there exists $d \in D$ such that $x = d\defon{n}$.
        Hence  $x \subseteq d\defon{n+1} \in T_D$.
        Thus $x$ is not a leaf, i.e.~$T_D$ is pruned.
    \item For any $T \subseteq  A^{<\omega}$, $[T]$ is a closed subset
        of $A^{\omega}$:

        Let $a \in A^{\omega} \setminus [T]$.
        Then there exists some $n$ such that $a\defon{n} \not\in T$.
        Hence $\{a_0\} \times \ldots \times \{a_{n+1}\} \times A^{\omega}$
        is an open neighbourhood of $a$ disjoint from $[T]$.

    \item $T \mapsto [T]$ is a bijection between the
        pruned trees on $A$ and the closed subsets of $A^{\omega}$.

        \begin{claim}
            $[T_D] = D$ for any closed subset $D \subseteq A^{\omega}$.
        \end{claim}
        \begin{subproof}
            Clearly $D \subseteq [T_D]$.
            Let $x \in [T_D]$.
            Then for every $n < \omega$,
            there exists some $d_n \in D$ such that
            $d_n\defon{n} = x\defon{n}$.
            Clearly the $d_n$ converge to $x$.
            Since $D$ is closed, we get $x \in D$.
        \end{subproof}

        This shows that $T \mapsto [T]$ is surjective.

        Now let $T \neq T'$ be pruned trees.
        Then there exists $x \in T \mathop{\triangle} T'$,
        wlog.~$x \in T \setminus T'$.
        Since $T$ is pruned
        by applying the axiom of countable choice
        we get an infinite branch $x' \in  [T] \setminus [T']$.
        Hence the map is injective.

    \item Let $N_s \coloneqq  \{x \in A^{\omega} | s \subseteq x\}$.
        Show that every open $U \subseteq A^{\omega}$
        can be written as $U = \bigcup_{s \in S}  N_s$
        for some set of pairwise incompatible $S \subseteq A^{<\omega}$.


        Let $U$ be open.
        Then $U$ has the form
        \[
        U = \bigcup_{i \in I} X_i \times  A^{\omega}
        \]
        for some $X_i \subseteq A^{n_i}$, $n_i < \omega$.
        Clearly
        $U = \bigcup_{s \in S'} N_s$
        for
        $S' \coloneqq  \bigcup_{i \in I} X_i$.
        Define
        \[
        S \coloneqq \{s \in S' | \lnot\exists t \in S'.~t\subseteq s \land |t| < |s|\}.
        \]
        Then the elements of $S$ are pairwise incompatible
        and $U = \bigcup_{s \in S}  N_s$.

    \item Let $T \subseteq A^{<\omega}$ be an infinite tree which is finitely
        splitting.
        Then $[T]$ is nonempty:

        Let us recursively construct a sequence of compatible $s_n \in T$
        with $|s_n| = n$
        such that  $\{s_n\} \times A^{<\omega} \cap T$ is infinite.
        Let  $s_0$ be the empty sequence;
        by assumption $T$ is infinite.
        Suppose that $s_n$ has been chosen.
        Since $T$ is finitely splitting, there are only finitely
        many $a \in A$ with $s_n\concat a \in T$.
        Since $ \{s_n\} \times A^{<\omega} \cap T$
        is infinite,
        there must exist at least on $a \in A$
        such that $\{s_n\concat a\} \times A^{<\omega} \cap T$
        is infinite.
        Define $s_{n+1} \coloneqq  s_n \concat a$.

        Then the union of the $s_n$ is an infinite branch of $T$,
        i.e.~$[T]$ is nonempty.


    \item Then $[T]$ is compact:
        \todo{TODO}

        % https://alanmath.wordpress.com/2011/06/16/on-trees-compactness-and-finite-splitting/
\end{enumerate}

\nr 2
\todo{handwritten}
\nr 3
\todo{handwritten}

\nr 4

\begin{notation}
    For $A \subseteq X$ let $A'$ denote the set of
    accumulation points of $A$.
\end{notation}

\begin{theorem}
    Let $X$ be a Polish space.
    Then there exists a unique partition $X = P \sqcup U$
    of $X$ into a perfect closed subset $P$ and a countable open subset $U$.
\end{theorem}
\begin{proof}

    Let $P$ be the set of condensation points of $X$
    and $U \coloneqq X \setminus P$.

    \begin{claim}
        $U$ is open and countable.
    \end{claim}
    \begin{subproof}
        Let $S$ be a countable dense subset.
        For each $x \in U$,
        there is an $\epsilon_x > 0$, $s_x \in S$ such that $x \in B_{\epsilon_x}(s_x)$
        is at most countable.
        Clearly $B_{\epsilon_x}(s_x) \subseteq U$,
        as for every $y \in B_{\epsilon_x}(s_x)$,
        $B_{\epsilon_x}(s_x)$ witnesses that $y \not\in P$.
        Thus $U = \bigcup_{x \in U} B_{\epsilon_x}(s_x)$
        is open.
        Wlog.~$\epsilon_x \in \Q$ for all $x$.
        Then the RHS is the union of at most
        countably many countable sets, as $S \times \Q$
        is countable.
    \end{subproof}

    \begin{claim}
        $P$ is perfect.
    \end{claim}
    \begin{subproof}
        Let $x \in P'$ and $x \in U$ an open neighbourhood.
        Then there exists $y \in P \cap U$.
        In particular, $U$ is an open neighbourhood of $ y$,
        hence $U$ is uncountable.
        It follows that $x \in P$.

        On the other hand let $x \in P$
        and let $U$ be an open neighbourhood.
        We need to show that $U \cap P \setminus \{x\}$
        is not empty.
        Suppose that for all $y \in U \cap P \setminus \{x\}$,
        there is an open neighbourhood $U_y$
        such that $U_y$ is at most countable.
        Wlog.~$U_y = B_{\epsilon_y}(s_y)$ for some $s_y \in S$, $\epsilon_y > 0$,
        where $S$ is again a countable dense subset.
        Wlog.~$\epsilon_y \in \Q$.
        But then
        \[
            U = \{x\} \cup \bigcup_{y \in U} B_{\epsilon_y}(s_y)
        \]
        is at most countable as a countable union of countable sets,
        contradiction $x \in P$.
    \end{subproof}

    \begin{claim}
        Let $P,U$ be defined as above
        and let $P_2 \subseteq  X$, $U_2 \subseteq  X$
        be such that $P_2$ is perfect and closed,
        $U_2$ is countable and open
        and $X = P_2 \sqcup U_2$.
        Then $P_2 = P$ and $U_2 = U$.
    \end{claim}
    \todo{TODO}
\end{proof}

\begin{corollary}\label{cor:polishcard}
    Any Polish space is either countable or has cardinality equal to $\fc$.
\end{corollary}
\begin{subproof}
    Let $X = P \sqcup U$
    where  $P$ is perfect and $U$ is countable.
    If $P \neq \emptyset$, we have $|P| = \fc$
    by \yaref{cor:perfectpolishcard}.
\end{subproof}
