Josia Pietsch
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8.1 KiB
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279 lines
8.1 KiB
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\lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}
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\subsection{Trees}
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\gist{%
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\begin{notation}
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Let $A \neq \emptyset$, $n \in \N$.
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Then
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\[
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A^n \coloneqq \{s\colon \{0,1,\ldots, n-1\} \to A \}
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\]
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is the set of $n$-element \vocab[Sequence]{sequences}.
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We often write
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$(s_0,s_1,\ldots,s_{n-1})$.
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If $s = (s_0,\ldots,s_{n-1})$,
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then $n$ is the \vocab{length} of $s$,
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denoted by $|s|$.
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If $n = 0$ there exists only the empty sequence,
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i.e.~$A^0 = \{\emptyset\}$ and $|\emptyset| = 0$.
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We set
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\[
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A^{<\N} \coloneqq \bigcup_{n=0}^{\infty} A^n
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\]
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and
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\[
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A^{\N} \coloneqq \{x \colon \N \to A\}.
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\]
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If $s \in A^n$ and $m \le n$,
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we let
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$s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$.
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Let $s,t \in A^{<\N}$.
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We say that $s$ is an \vocab{initial segment}
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of $t$ (or $t$ is an \vocab{extension} of $s$)
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if there exists an $n$ such that $s = t\defon{|s|}$.
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We write this as $s \subseteq t$.
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We say that $s$ and $t$ are \vocab{compatible}
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if $s \subseteq t$ or $t \subseteq s$.
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Otherwise the are \vocab{incompatible},
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we denote that as $s \perp t$.
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The \vocab{concatenation}
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of $s = (s_0,\ldots, s_{n-1})$
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and $t = (t_0,\ldots, t_{m-1})$
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is the sequence
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$s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$
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In the case of $t = (a)$
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we also write $s\concat a$ for $s\concat (a)$.
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Similarly, if $x \in A^{\N}$
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we can write $x = (x_0,x_1,\ldots)$.
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If $n \in \N$, $x\defon{n} \coloneqq (x_0,\ldots,x_{n-1})$,
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define extension, initial segments
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and concatenation of a finite sequence with an infinite one.
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\end{notation}
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\begin{definition}
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A \vocab{tree}
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on a set $A$ is a subset $T \subseteq A^{<\N}$
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closed under initial segments,
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i.e.~if $t \in T, s \subseteq t \implies s \in T$.
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Elements of trees are called \vocab{nodes}.
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A \vocab{leave} is an element of $T$,
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that has no extension in $t$.
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An \vocab{infinite branch} of a tree $T$
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is $x \in A^{\N}$
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such that $\forall n.~x\defon{n} \in T$.
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The \vocab{body} of $T$ is the set of all
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infinite branches:
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\[
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[T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.
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\]
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We say that $T$ is \vocab{pruned},
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iff
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\[
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\forall t\in T.\exists s \supsetneq t.~s \in T.
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\]
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\end{definition}
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}{}
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\begin{definition}
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A \vocab{Cantor scheme}
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on a set $X$ is a family
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$(A_s)_{s \in 2^{< \N}}$
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of subsets of $X$ such that
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\begin{enumerate}[i)]
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\item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
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\item $\forall s \in 2^{<\N}, i \in 2.~A_{s \concat i} \subseteq A_s$.
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\end{enumerate}
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\end{definition}
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\begin{definition}
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A topological space
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is \vocab{perfect}
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if it has no isolated points,
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i.e.~for any $U \neq \emptyset$ open,
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there $x \neq y$ such that $x, y \in U$.
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\end{definition}
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\begin{theorem}
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\label{thm:cantortopolish}
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Let $X \neq \emptyset$
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be a perfect Polish space.
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Then there is an embedding
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of the Cantor space $2^{\N}$
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into $X$.
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\end{theorem}
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\begin{proof}
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We will define a Cantor scheme
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$(U_s)_{s \in 2^{<\N}}$
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such that $\forall s \in 2^{< \N}$.
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\begin{enumerate}[(i)]
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\item $U_s \neq \emptyset$ and open,
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\item $\diam(U_s) \le 2^{-|s|}$,
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\item $\overline{U_{s \concat i}} \subseteq U_s$
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for $i \in 2$.
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\end{enumerate}
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We define $U_s$ inductively on the length of $s$.
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\gist{%
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For $U_{\emptyset}$ take any non-empty open set
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with small enough diameter.
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Given $U_s$, pick $x \neq y \in U_s$
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and let $U_{s \concat 0} \ni x$,
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$U_{s \concat 1} \ni y$
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be disjoint, open,
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of diameter $\le \frac{1}{2^{|s| +1}}$
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and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
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}{}
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\gist{%
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Let $x \in 2^{\N}$.
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Then let $f(x)$ be the unique point in $X$
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such that
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\[
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\{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}.
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\]
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(This is nonempty as $X$ is a completely metrizable space.)
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It is clear that $f$ is injective and continuous.
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% TODO: more details
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$2^{\N}$ is compact, hence $f^{-1}$ is also continuous.
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}{Consider $f\colon 2^{\N} \hookrightarrow X, x \mapsto y$, where $\{y\} = \bigcap_n U_{x\defon n}$.
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By compactness of $2^{\N}$, we get that $f^{-1}$ is continuous.}
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\end{proof}
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\begin{corollary}
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\label{cor:perfectpolishcard}
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Every nonempty perfect Polish
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space $X$ has cardinality $\fc = 2^{\aleph_0}$
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\end{corollary}
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\begin{proof}
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\gist{%
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Since the cantor space embeds into $X$,
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we get the lower bound.
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Since $X$ is second countable and Hausdorff,
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we get the upper bound:
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Let $\langle U_n : n < \omega \rangle$ be a countable basis.
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Consider the injective function
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\begin{IEEEeqnarray*}{rCl}
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f\colon X &\longrightarrow & 2^{ \omega} \\
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x &\longmapsto & \{n : x \in U_n\}.
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\end{IEEEeqnarray*}
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}{Lower bound: $2^{\N} \hookrightarrow X$,
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upper bound: \nth{2} countable and Hausdorff.}
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\end{proof}
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\begin{theorem}
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Any Polish space is countable
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or it has cardinality $\fc$. % TODO C
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\end{theorem}
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\begin{proof}
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See \autoref{cor:polishcard}.
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\end{proof}
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\begin{definition}
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A \vocab{Lusin scheme} on a set $X$
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is a family $(A_s)_{s \in \N^{<\N}}$
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of subsets of $X $
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such that
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\begin{enumerate}[(i)]
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\item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$
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for all $j \neq i \in \N$, $s \in \N^{<\N}$.
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\item $A_{s \concat i} \subseteq A_s$
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for all $i \in \N, s \in \N^{<\N}$.
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\end{enumerate}
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\end{definition}
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\begin{theorem}
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\label{thm:bairetopolish}
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Let $X \neq \emptyset$ be a Polish space.
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Then there is a closed subset
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\[
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D \subseteq \N^\N \mathbin{\text{\reflectbox{$\coloneqq$}}} \cN
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\]
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and a continuous bijection $f\colon D \to X$
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(the inverse does not need to be continuous).
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Moreover there is a continuous surjection $g: \cN \to X$
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extending $f$.
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\end{theorem}
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\begin{definition}
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An $F_\sigma$ set is the
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countable union of closed sets,
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i.e.~the complement of a $G_\delta$ set.
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\end{definition}
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\gist{%
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\begin{observe}
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\begin{itemize}
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\item Any open set is $F {\sigma}$.
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\item In metric spaces the intersection of an open and closed set is $F_\sigma$.
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\end{itemize}
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\end{observe}
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}{}
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\begin{refproof}{thm:bairetopolish}
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Let $d$ be a complete metric on $X$.
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W.l.o.g.~$\diam(X) \le 1$.
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We construct a Lusin scheme
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$(F_s)_{s \in \N^{<\N}}$
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such that $F_s \subseteq X$
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and
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\begin{enumerate}[(i)]
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\item $F_\emptyset = X$,
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\item $F_s$ is $F_\sigma$ for all $s$.
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\item The $F_{s \concat i}$ partition $F_s$,
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i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.
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Furthermore we want that
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$\overline{F_{s \concat i}} \subseteq F_s$
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for all $i$.
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\item $\diam(F_s) \le 2^{-|s|}$.
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\end{enumerate}
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\gist{%
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Suppose we already have $F_s \mathbin{\text{\reflectbox{$\coloneqq$}}} F$.
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We need to construct a partition $(F_i)_{i \in \N}$
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of $F$ with $\overline{F_i} \subseteq F$
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and $\diam(F_i) < \epsilon$
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for $\epsilon = 2^{-|s| - 1}$,
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such that the $F_i$ are $F_\sigma$.
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\paragraph{Step 1}
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Write $F \coloneqq \bigcup_{i \in \N} C_i$
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for some closed sets $C_i$.
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W.l.o.g.~$C_i \subseteq C_{i+1}$.
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Let $F_i^0 \coloneqq C_{i+1} \setminus C_i$.
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These $F_i^0$ are $F_\sigma$,
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and form a partition of $F$.
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Furthermore $\overline{F_i^0} \subseteq F$.
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However the diameter might be too large.
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Fix $i \in \N$ and consider $F_i^0$.
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Cover it with countably many open balls $B_1, B_2,\ldots$
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of diameter smaller than $\epsilon$.
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The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
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are $F_\sigma$, disjoint
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and $F_i^0 = \bigcup_{j} D_j$.
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}{Induction.}
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\phantom\qedhere
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\end{refproof}
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