\lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}

\subsection{Trees}

\gist{%
\begin{notation}
    Let $A \neq  \emptyset$, $n \in \N$.
    Then
    \[
        A^n \coloneqq  \{s\colon \{0,1,\ldots, n-1\} \to A \}
    \]
    is the set of $n$-element \vocab[Sequence]{sequences}.
    We often write
    $(s_0,s_1,\ldots,s_{n-1})$.

    If $s = (s_0,\ldots,s_{n-1})$,
    then $n$ is the \vocab{length} of $s$,
    denoted by $|s|$.

    If $n = 0$ there exists only the empty sequence,
    i.e.~$A^0 = \{\emptyset\}$ and $|\emptyset| = 0$.

    We set
    \[
    A^{<\N} \coloneqq  \bigcup_{n=0}^{\infty} A^n
    \]
    and
    \[
    A^{\N} \coloneqq \{x \colon \N \to  A\}.
    \]

    If $s \in A^n$ and $m \le n$,
    we let
    $s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$.

    Let $s,t \in A^{<\N}$.
    We say that $s$ is an \vocab{initial segment}
    of $t$ (or $t$ is an \vocab{extension} of $s$)
    if there exists an $n$ such that $s = t\defon{|s|}$.
    We write this as $s \subseteq t$.

    We say that $s$ and $t$ are \vocab{compatible}
    if $s \subseteq t$ or $t \subseteq s$.
    Otherwise the are \vocab{incompatible},
    we denote that as $s \perp t$.

    The \vocab{concatenation}
    of $s = (s_0,\ldots, s_{n-1})$
    and $t = (t_0,\ldots, t_{m-1})$
    is the sequence
    $s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$

    In the case of $t = (a)$
    we also write $s\concat a$ for $s\concat (a)$.

    Similarly, if $x \in A^{\N}$
    we can write $x = (x_0,x_1,\ldots)$.
    If $n \in \N$, $x\defon{n} \coloneqq (x_0,\ldots,x_{n-1})$,
    define extension, initial segments
    and concatenation of a finite sequence with an infinite one.
\end{notation}

\begin{definition}
    A \vocab{tree}
    on a set $A$ is a subset $T \subseteq  A^{<\N}$
    closed under initial segments,
    i.e.~if $t \in T, s \subseteq t \implies s \in T$.
    Elements of trees are called \vocab{nodes}.

    A \vocab{leave} is an element of $T$,
    that has no extension in $t$.

    An \vocab{infinite branch} of a tree $T$
    is $x \in A^{\N}$
    such that $\forall n.~x\defon{n} \in T$.

    The \vocab{body}  of $T$ is the set of all
    infinite branches:
    \[
    [T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.
    \]

    We say that $T$ is \vocab{pruned},
    iff
    \[
    \forall  t\in T.\exists  s \supsetneq t.~s \in T.
    \]
\end{definition}
}{}

\begin{definition}
    A \vocab{Cantor scheme}
    on a set $X$ is a family
    $(A_s)_{s \in 2^{< \N}}$
    of subsets of $X$ such that
    \begin{enumerate}[i)]
        \item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
        \item $\forall s \in 2^{<\N}, i \in 2.~A_{s \concat i} \subseteq A_s$.
    \end{enumerate}
\end{definition}

\begin{definition}
    A topological space
    is \vocab{perfect}
    if it has no isolated points,
    i.e.~for any $U \neq \emptyset$ open,
    there $x \neq y$ such that $x, y \in U$.
\end{definition}

\begin{theorem}
    \label{thm:cantortopolish}
    Let $X \neq \emptyset$
    be a perfect Polish space.
    Then there is an embedding
    of the Cantor space $2^{\N}$
    into $X$.
\end{theorem}
\begin{proof}
    We will define a Cantor scheme
    $(U_s)_{s \in 2^{<\N}}$
    such that $\forall s \in  2^{< \N}$.
    \begin{enumerate}[(i)]
        \item $U_s \neq  \emptyset$ and open,
        \item $\diam(U_s) \le  2^{-|s|}$,
        \item $\overline{U_{s  \concat i}} \subseteq  U_s$
            for $i \in 2$.
    \end{enumerate}

    We define $U_s$ inductively on the length of $s$.

    \gist{%
        For $U_{\emptyset}$ take any non-empty open set
        with small enough diameter.

        Given $U_s$, pick $x \neq y \in U_s$
        and let $U_{s \concat 0} \ni x$,
        $U_{s \concat 1} \ni y$
        be disjoint, open,
        of diameter $\le \frac{1}{2^{|s| +1}}$
        and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq  U_s$.
    }{}

    \gist{%
    Let $x \in  2^{\N}$.
    Then let $f(x)$ be the unique point in $X$
    such that
    \[
    \{f(x)\}  = \bigcap_{n}  U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}.
    \]
    (This is nonempty as $X$ is a completely metrizable space.)
    It is clear that $f$ is injective and continuous.
    % TODO: more details
    $2^{\N}$ is compact, hence $f^{-1}$ is also continuous.
    }{Consider $f\colon 2^{\N} \hookrightarrow X, x \mapsto y$, where $\{y\}  = \bigcap_n U_{x\defon n}$.
    By compactness of $2^{\N}$, we get that  $f^{-1}$ is continuous.}
\end{proof}

\begin{corollary}
    \label{cor:perfectpolishcard}
    Every nonempty perfect Polish
    space $X$ has cardinality $\fc = 2^{\aleph_0}$
\end{corollary}
\begin{proof}
    \gist{%
        Since the cantor space embeds into $X$,
        we get the lower bound.
        Since $X$ is second countable and Hausdorff,
        we get the upper bound:
        Let $\langle U_n : n < \omega \rangle$ be a countable basis.
        Consider the injective function
        \begin{IEEEeqnarray*}{rCl}
            f\colon X &\longrightarrow & 2^{ \omega} \\
             x &\longmapsto & \{n : x \in U_n\}.
        \end{IEEEeqnarray*}
    }{Lower bound:  $2^{\N} \hookrightarrow X$,
        upper bound: \nth{2} countable and Hausdorff.}
\end{proof}

\begin{theorem}
    Any Polish space is countable
    or it has cardinality $\fc$. % TODO C
\end{theorem}
\begin{proof}
    See \autoref{cor:polishcard}.
\end{proof}


\begin{definition}
    A \vocab{Lusin scheme} on a set $X$
    is a family  $(A_s)_{s \in \N^{<\N}}$
    of subsets of $X $
    such that
    \begin{enumerate}[(i)]
        \item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$
            for all $j \neq i \in \N$, $s \in \N^{<\N}$.
        \item $A_{s \concat i} \subseteq  A_s$
            for all $i \in \N, s \in \N^{<\N}$.
    \end{enumerate}
\end{definition}

\begin{theorem}
    \label{thm:bairetopolish}
    Let $X \neq \emptyset$ be a Polish space.
    Then there is a closed subset
    \[
        D \subseteq \N^\N \mathbin{\text{\reflectbox{$\coloneqq$}}} \cN
    \]
    and a continuous bijection $f\colon D \to X$
    (the inverse does not need to be continuous).

    Moreover there is a continuous surjection $g: \cN \to  X$
    extending $f$.
\end{theorem}
\begin{definition}
    An $F_\sigma$ set is the
    countable union of closed sets,
    i.e.~the complement of a $G_\delta$ set.
\end{definition}
\gist{%
\begin{observe}
    \begin{itemize}
        \item Any open set is $F {\sigma}$.
        \item In metric spaces the intersection of an open and closed set is $F_\sigma$.
    \end{itemize}
\end{observe}
}{}
\begin{refproof}{thm:bairetopolish}
    Let $d$ be a complete metric on $X$.
    W.l.o.g.~$\diam(X) \le 1$.
    We construct a Lusin scheme
    $(F_s)_{s \in \N^{<\N}}$
    such that $F_s \subseteq  X$
    and
    \begin{enumerate}[(i)]
        \item $F_\emptyset = X$,
        \item $F_s$ is $F_\sigma$ for all  $s$.
        \item The $F_{s \concat i}$ partition $F_s$,
            i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.

            Furthermore we want that
            $\overline{F_{s \concat i}} \subseteq F_s$
            for all $i$.
        \item $\diam(F_s) \le 2^{-|s|}$.
    \end{enumerate}

    \gist{%
    Suppose we already have $F_s \mathbin{\text{\reflectbox{$\coloneqq$}}} F$.
    We need to construct a partition $(F_i)_{i \in \N}$
    of $F$ with $\overline{F_i} \subseteq F$
    and $\diam(F_i) < \epsilon$
    for $\epsilon = 2^{-|s| - 1}$,
    such that the $F_i$ are $F_\sigma$.

    \paragraph{Step 1}
    Write $F \coloneqq \bigcup_{i \in \N} C_i$
    for some closed sets $C_i$.
    W.l.o.g.~$C_i \subseteq C_{i+1}$.

    Let $F_i^0 \coloneqq C_{i+1} \setminus C_i$.
    These $F_i^0$ are $F_\sigma$,
    and form a partition of $F$.
    Furthermore $\overline{F_i^0} \subseteq F$.

    However the diameter might be too large.
    Fix $i \in \N$ and consider $F_i^0$.
    Cover it with countably many open balls  $B_1, B_2,\ldots$
    of diameter smaller than  $\epsilon$.
    The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
    are $F_\sigma$, disjoint
    and $F_i^0 = \bigcup_{j} D_j$.
}{Induction.}



    \phantom\qedhere
\end{refproof}


