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\lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}
Let $A \neq \emptyset$, $n \in \N$.
A^n \coloneqq \{s\colon \{0,1,\ldots, n-1\} \to A \}
is the set of $n$-element \vocab[Sequence]{sequences}.
We often write
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
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.
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},
\forall t\in T.\exists s \supsetneq t.~s \in T.
A \vocab{Cantor scheme}
on a set $X$ is a family
$(A_s)_{s \in 2^{< \N}}$
of subsets of $X$ such that
\item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
\item $\forall s \in 2^{<\N}, i \in 2.~A_{s \concat i} \subseteq A_s$.
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 $x, y \in U$.
Let $X \neq \emptyset$
be a perfect Polish space.
Then there is an embedding
of the Cantor space $2^{\N}$
into $X$.
We will define a Cantor scheme
$(U_s)_{s \in 2^{<\N}}$
such that $\forall s \in 2^{< \N}$.
\item $U_s \neq \emptyset$ and open,
\item $\diam(U_s) \le 2^{-|s|}$,
\item $\overline{U_{s \concat i}} \subseteq U_s$
for $i \in 2$.
We define $U_s$ inductively on the length of $s$.
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 $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
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.}
Every nonempty perfect Polish
space $X$ has cardinality $\fc = 2^{\aleph_0}$
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
f\colon X &\longrightarrow & 2^{ \omega} \\
x &\longmapsto & \{n : x \in U_n\}.
}{Lower bound: $2^{\N} \hookrightarrow X$,
upper bound: \nth{2} countable and Hausdorff.}
Any Polish space is countable
or it has cardinality $\fc$. % TODO C
See \autoref{cor:polishcard}.
A \vocab{Lusin scheme} on a set $X$
is a family $(A_s)_{s \in \N^{<\N}}$
of subsets of $X $
such that
\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 $i \in \N, s \in \N^{<\N}$.
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 $f$.
An $F_\sigma$ set is the
countable union of closed sets,
i.e.~the complement of a $G_\delta$ set.
\item Any open set is $F {\sigma}$.
\item In metric spaces the intersection of an open and closed set is $F_\sigma$.
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 $F_s \subseteq X$
\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 $\diam(F_s) \le 2^{-|s|}$.
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 $F_i^0 = \bigcup_{j} D_j$.