Josia Pietsch 324dddda04
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Sheet 9, exercise 4
2024-01-06 19:04:51 +01:00

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\tutorial{10}{2023-12-19}{Sheet 9}
\subsection{Sheet 9}
\nr 1
$(X, \tau') \xrightarrow{x \mapsto x} (X, \tau)$ is Borel
(by one of the equivalent definitions of being Borel).
Thus $\cB(X, \tau) \subseteq \cB(X, \tau')$
(by the other equivalent definition of being Borel).
Let $U \subseteq (X, \tau')$ be Borel.
$\id\defon{U}$ is injective,
hence $U$ is Borel in $(X, \tau)$ by Lusin-Suslin.
\paragraph{Related stuff}
Let $X,Y$ be Polish.
$f\colon X \to Y$
is Borel iff its graph $\Gamma_f$ is Borel.
Take a countable open base
$V_0, V_1, \ldots$ of $Y$.
Then $\Gamma_f = \{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}$
(because the space is Hausdorff).
If $f$ is Borel,
then clearly the RHS is Borel
&&\{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}\\
&=& \bigcap_{n < \omega} (f^{-1}(V_n)^{c}Y \cup f^{-1}(V_n) \times V_n\}\\
On the other hand suppose that $\Gamma_f$ is Borel.
f^{-1}(B) = \pi_X(X \times B \cap \Gamma_f)
is analytic.\footnote{Note that the projection of a Borel set is not necessarily Borel.
Moreover note that we only used that $\Gamma_f$ is analytic.}
On the other hand
f^{-1}(B)^c = f^{-1}(B^c)
is analytic
and we know that $\Sigma_1^1 \cap \Pi_1^1 = \cB$
by the \yaref{cor:lusinseparation}.
In fact we have shown
The following are equivalent
\item $f$ is Borel,
\item $\Gamma_f$ is Borel,
\item $\Gamma_f$ is analytic.
\nr 2
Let $X$ be a topological space.
Let $K(X)$ be the set of all compact subspaces of $X$.
The \vocab{Vietoris Topology}, $\tau_V$, on $K(X)$
is the topology with basic open sets
[U_0; U_1, \ldots, U_n] = \{K \in K(X) : K \subseteq U_0 \land \forall 1 \le i \le n .~K \cap U_i \neq \emptyset\}
for $U_i \overset{\text{open}}{\subseteq} X$.
Let $(X,d)$ be a matric space with $d \le 1$.
We define a metric $d_H$ on $K(X)$ as follows:
$d_H(\emptyset, \emptyset) \coloneqq 0$, $d_H(K, \emptyset) \coloneqq 1$ for $K \neq \emptyset$
d_H(K_0, K_1) \coloneqq \max \{\max_{x \in K_0} d(x,K_1), \max_{x \in K_1} d(x,K_0)\}
for $K_0, K_1 \neq \emptyset$.
$d_H$ is indeed a metric.
Let $\delta(K, L) \coloneqq \max_{x \in K} d(x,L)$.
It suffices to show
$\delta(X,Z) \le \delta(X,Y) + \delta(Y,Z)$,
since then
d_H(X,Z) &\le & \max \{\delta(X,Y) + \delta(Y,Z), \delta(Z,Y) + \delta(Y,X)\}\\
&\le & d_H(X,Y) + d_H(Y,Z).
Using the fact that $d(\cdot , Z)$ is uniformly continuous,
|d(x,Z) - d(y,Z)| \le d(x,y), % TODO REF SHEET 1
we get
&d(x,Z) &\le & d(x,y) + d(y, Z)\\
& &\le & d(x,y) + \delta(Y,Z)\\
\implies& d(x,Z) - \delta(Y,Z) &\le & d(x,Y)\\
\implies& d(x,Z) &\le & \delta(X,Y) + \delta(Y,Z)\\
\implies & \delta(X,Z) &\le & \delta(X,Y) + \delta(Y,Z).
\item We have
d_H(K_0, K_1) < \epsilon & \iff & \max \{\max_{x \in K_0}d(x, K_1), \max_{x \in K_1} d(x,K_0)\} < \epsilon\\
&\iff& \max_{x \in K_0} d(x, K_1) < \epsilon \land \max_{x \in K_1} d(x, K_0) < \epsilon\\
&\iff& K_0 \subseteq B_{\epsilon}(K_1) \land K_1 \subseteq B_\epsilon(K_0).
\item Note that a subbase of $\tau_V$
is given by $[U]$ and $\langle U \rangle \coloneqq [X;U]$ for $U \overset{\text{open}}{\subseteq} X$.
Let $K \in [U]$.
Then $d(\cdot , U^c)\colon U \to \R_{\ge 0}$
is always non-zero and continuous.
So $d(K,U^c)$ attains a minimum $\epsilon > 0$.
Then $B_{\epsilon}^H(K) \subseteq U$,
so $[U]$ is open in $\tau_V$.
Let $K \in \langle U \rangle$.
Take some $k \in K \cap U$.
Then there is some $\epsilon > 0$
such that $B_\epsilon(k) \subseteq U$.
Then $K \in B_{\epsilon}^H(K) \subseteq \langle U \rangle$.
\todo{Other direction}
% $\tau_H \subseteq \tau_V$
\item Consider a countable dense subset of $X$.
Let $\cK$ be the set of finite subsets of that countable dense subset.
Then $\cK \subseteq K(X)$ is dense:
Take $K \in K(X)$ an let $\epsilon > 0$.
$K$ can be covered with finitely many $\epsilon$-balls
with centers from the countable dense subsets.
Let $K' \in \cK$ be the set of the centers.
Then $d_H(K, K') \le \epsilon$.
\nr 3
\item By transfinite induction we get that $\alpha$ is an ordinal,
since $\prec$ is well-founded and the supremum of a sets
of ordinals is an ordinal.
Since $\rho_{\prec}\colon X \to \alpha$
is a surjection, it follows that $\alpha \le |X|$,
i.e.~$\alpha < |X|^+$.
\item By induction on $\rho_{\prec_X}(x)$ we show that
$\rho_{\prec_{X}}(x) \le \rho_{\prec_Y}(f(x))$.
For $0$ this is trivial.
Suppose that $\rho_{\prec_{X}}(x) = \alpha$
and the statement was shown for all $\beta < \alpha$.
\rho_{\prec_Y}(f(x)) &=& \sup \{\rho_{\prec_Y}(y') + 1 | y' \prec f(x)\}\\
&\ge& \sup \{\rho_{\prec_Y}(f(x')) + 1 | f(x') \prec f(x)\}\\
&\ge & \sup \{\rho_{\prec_Y}(f(x')) + 1 | x' \prec x\}\\
&\ge & \sup \{\rho_{\prec_X}(x') + 1 | x' \prec x\}\\
&=& \rho_{\prec_X}(x).
\item Infinite branches of $T_\prec$ correspond
to infinite descending chain of $\prec$,
hence $T_{\prec}$ is well-founded iff $\prec$ is well-founded.
% Unwarp$^{\text{tm}}$ definitions.
Suppose that $\prec$ is well-founded.
Note that $\rho_T(s)$ depends only on the last element of $s$,
as for $s, s' \in T$ with the same last element,
we have $s \concat x \in T \iff s' \concat x \in T$.
Let $s = (s_0, \ldots, s_n)$.
Let us show that $\rho_T(s) = \rho_{\prec}(s_n)$.
We use induction on $\rho_T(s)$.
For leaves this is immediate.
From the last exercise sheet we know that
\rho_T(s) = \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}.
\rho_T(s) &=& \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}\\
&=& \sup \{\rho_{\prec}(a) + 1 | s \concat a \in T\}\\
&=& \sup \{\rho_{\prec}(a) + 1 | a \prec s_n\}\\
&=& \rho_\prec(s_n).
\nr 4
A solution can be found in \cite{coanalyticranks}.
% TODO Copy relevant points