\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 from $D$ onto $X$ (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}