Josia Pietsch 3df55b6516
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Note that $B_{(n_0,\ldots, n_k)} \subseteq B_{(n_0,\ldots, n_k)}^\ast \subseteq \overline{B_{n_0,\ldots, n_k}}$.
We want to show that
f(A) = \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast.
Let $x \in f(A)$.
Then take $a \in A$ such that $x = f(a)$.
\[x \in \bigcap_k \underbrace{B_{a\defon{k}}}_{= f(A \cap N_{a\defon{k}})} \subseteq \bigcap_{k} B^\ast_{a\defon{k}}.\]
This gives $f(A) \subseteq \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast$.
If $x \in \bigcap_{k < \omega} \bigcup_{s \in \omega^k} B_s^\ast$,
Then there is a unique $a$ such that
$x \in \bigcap_k B^\ast_{a\defon{k}}$.
$a \in A$.
We have $B^\ast_{a\defon{k}} \subseteq \overline{B_{a\defon{k}}}$.
So $x \in \bigcap_k \overline{B_{a\defon{k}}}$.
In particular, $B_{a\defon{k}} \neq \emptyset$
for all $k$.
So for all $k$ we get that $A \cap N_{a\defon{k}} \neq \emptyset$.
But $A$ is closed and $N_{a\defon{k}}$
is clopen for all $k$.
We have $\{a\} = \bigcup_k N_{a\defon{k}}$,
so $a \in A$.
$f(a) = x$.
We have $f(a) \in \bigcap_k B_{a\defon{k}}$.
Suppose $f(a) \neq x$.
Pick $U \ni f(a)$ open
such that $x \not\in \overline{U}$.
By continuity of $f$,
we get that $f(N_{a\defon{k_0}}) \subseteq U$
for $k_0$ large enough.
So $x \not\in \overline{f(N_{a\defon{k_0}})}$.
In particular
$x \not\in \overline{f(N_{a\defon{k_0}})} = \overline{B_{a\defon{k_0}}} \supset B^\ast_{a\defon{k_0}}$.
But $x \in \bigcap_k B^\ast_{a\defon{k}} \lightning$.
\begin{corollary}[of the \yaref{thm:lusinseparation}]
\yalabel{Corollary of the Lusin Separation Theorem}{Lusin Separation}{cor:lusinseparation}
Let $X$ be Polish.
Let $A_1, A_2, A_3,\ldots \subseteq X$ be analytic
and pairwise disjoint.
Then there are pairwise disjoint Borel sets $B_i \supseteq A_i$.
For all $i$, let $B_i, C_i$
be disjoint Borel sets,
such that $A_i \subseteq B_i$
and $\bigcup_{j \neq i} A_j \subseteq C_i$.
Take $D_i \coloneqq B_i \cap \bigcap_{j \neq i} C_j$.
\begin{theorem}[\vocab{Borel Schröder-Bernstein}]
\yalabel{Schröder-Bernstein for Borel sets}{Schröder-Bernstein}{thm:bsb}
Let $A, B$ be Borel in some Polish spaces.
Suppose that there are Borel embeddings
$f\colon A \hookrightarrow B$
and $g\colon B \hookrightarrow A$.
Then $A$ and $B$ are Borel isomorphic.
\begin{theorem}[\vocab{Isomorphism Theorem}]
\yalabel{Isomorphism Theorem}{Isomorphism Thm.}{thm:isomorphism}
Let $X, Y$ be Borel in some Polish spaces.
Then $X$ is Borel isomorphic
to $Y$ iff $|X| = |Y|$.
$\implies$ is clear.
Suppose that $|X| = |Y| \le \aleph_0$,
then any bijection suffices,
since all subsets are Borel.
If $|X| = |Y| > \aleph_0$,
then they must have cardinality $\fc$,
since we can embed the Cantor space.
It suffices to show that if $X$ is an uncountable
Polish space and $\cC = 2^\omega$ the Cantor space,
then they are Borel isomorphic.
There is $2^\omega \hookrightarrow X$ Borel
(continuous wrt.~the topology of $X$)
On the other hand
X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC
For the first inclusion,
recall that there is a continuous bijection $b\colon D \to X$,
where $D \overset{\text{closed}}{\subseteq} \cN$.
Consider $b^{-1}$.
Whenever $B \subseteq X$ is Borel,
we have that $b^{-1}(B)$ is Borel,
since $b$ is continuous.
For $A \subseteq D$ Borel
be get by \yaref{thm:lusinsouslin},
that $b$ with respect to $b(A)$
is Borel,
since $b\defon{A}$ is injective.
Hence \yaref{thm:bsb} can be applied.
\subsection{The Projective Hierarchy}
% https://q.uiver.app/#q=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
& {\Sigma^1_1(X)} && {\Sigma^1_2(X)} \\
{\Delta^1_1(X)} && {\Delta^1_2(X)} \\
& {\Pi^1_1(X)} && {\Pi^1_2(X)}
\arrow["\subseteq", hook, from=2-1, to=1-2]
\arrow["\subseteq"', hook, from=2-1, to=3-2]
\arrow["\subseteq"', hook, from=3-2, to=2-3]
\arrow["\subseteq", hook, from=1-2, to=2-3]
\arrow["\subseteq", hook, from=2-3, to=1-4]
\arrow["\subseteq", hook, from=2-3, to=3-4]
Let $X$ be a Polish space.
We define
\Delta^1_n(X) &\coloneqq& \Sigma^1_n(X) \cap \Pi^1_n(X)\\
\Pi^1_n(X) &=& \{A \subseteq X : X \setminus A \in \Sigma^1_n(X)\}\\
\Sigma^1_{n+1}(X) &=& \{ A \subseteq X : \exists B \in \Pi^1_n(X \times \cN) .~A = \proj_X[B]\}
Every analytic and every coanalytic set
has the Baire property.
We will not proof this in this lecture.
\subsection{Ill-Founded Trees}
Recall that a \vocab{tree} on $\N$ is a subset of
closed under taking initial segments.
We now identify trees with their characteristic functions,
i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}%
{We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:}
\One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\
x &\longmapsto & \begin{cases}
1 &: x \in T,\\
0 &: x \not\in T.
\gist{Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.}{}
Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
$\Tr \subseteq {2^{\N}}^{<\N}$ is closed
(where we take the topology of the Cantor space).
Indeed, for any $ s \in \N^{<\N}$
we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
Boolean combinations of such sets are clopen as well.
In particular for $s$ fixed,
we have that
\[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
is clopen in ${2^{\N}}^{<\N}$.