This commit is contained in:
parent
d001b56d71
commit
bfcb894c89
GPG Key ID:
E70B571D66986A2D
192
inputs/lecture_07.tex
Normal file
192
inputs/lecture_07.tex
Normal file
@ -0,0 +1,192 @@
|
||||
\lecture{07}{2023-11-07}{}
|
||||
|
||||
\begin{proposition}
|
||||
Let $X$ be second countable.
|
||||
Then $|\cB(X)| \le \fc$.
|
||||
% $\fc := 2^{\aleph_0}$
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
We use strong induction on $\xi < \omega_1$.
|
||||
We have $\Sigma^0_1(X) \le \fc$
|
||||
(for every element of the basis, we can decide
|
||||
whether to use it in the union or not).
|
||||
|
||||
Suppose that $\forall \xi' < \xi.~|\Sigma^0_{\xi'}(X)| \le \fc$.
|
||||
Then $|\Pi^0_{\xi'}(X)| \le \fc$.
|
||||
We have that
|
||||
\[
|
||||
\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
|
||||
\]
|
||||
Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.
|
||||
|
||||
We have
|
||||
\[
|
||||
\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).
|
||||
\]
|
||||
Hence
|
||||
\[
|
||||
|\cB(X)| \le \omega_1 \cdot \fc = \fc.
|
||||
\]
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}[Closure properties]
|
||||
Suppose that $X$ is metrizable.
|
||||
Let $1 \le \xi < \omega_1$.
|
||||
Then
|
||||
\begin{enumerate}[(a)]
|
||||
\item \begin{itemize}
|
||||
\item $\Sigma^0_\xi(X)$ is closed under countable unions.
|
||||
\item $\Pi^0_\xi(X)$ is closed under countable intersections.
|
||||
\item $\Delta^0_\xi(X)$ is closed under complements,
|
||||
countable unions and
|
||||
countable intersections.
|
||||
\end{itemize}
|
||||
\item \begin{itemize}
|
||||
\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
|
||||
\item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
|
||||
\end{itemize}
|
||||
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
\begin{enumerate}[(a)]
|
||||
\item This follows directly from the definition.
|
||||
Note that a countable intersection can be written
|
||||
as a complement of the countable union of complements:
|
||||
\[
|
||||
\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
|
||||
\]
|
||||
\item If suffices to check this for $\Sigma^0_{\xi}(X)$.
|
||||
Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
|
||||
and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
|
||||
Then
|
||||
\[
|
||||
A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)
|
||||
\]
|
||||
and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
|
||||
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
\begin{example}
|
||||
Consider the cantor space $2^{\omega}$.
|
||||
We have that $\Delta^0_1(2^{\omega})$
|
||||
is not closed under countable unions
|
||||
(countable unions yield all open sets, but there are open
|
||||
sets that are not clopen).
|
||||
\end{example}
|
||||
|
||||
\subsection{Turning Borels Sets into Clopens}
|
||||
|
||||
\begin{theorem}
|
||||
\label{thm:clopenize}
|
||||
Let $(X, \cT)$ be a Polish space.
|
||||
For any Borel set $A \subseteq X$,
|
||||
there is a finer Polish topology,
|
||||
\footnote{i.e.~$\cT_A \supseteq \cT$ and $(X, \cT_A)$ is Polish}
|
||||
such that
|
||||
\begin{itemize}
|
||||
\item $A$ is clopen in $\cT_A$,
|
||||
\item the Borel sets do not change,
|
||||
i.e.~$\cB(X, \cT) = \cB(X, \cT_A)$.
|
||||
\end{itemize}
|
||||
\end{theorem}
|
||||
\begin{corollary}[Perfect set property]
|
||||
Let $(X, \cT)$ be Polish,
|
||||
and let $B \subseteq X$ be Borel and uncountable.
|
||||
Then there is an embedding
|
||||
of the cantor space $2^{\omega}$
|
||||
into $B$.
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
Pick $\cT_B \supset \cT$
|
||||
such that $(X, \cT_B)$ is Polish,
|
||||
$B$ is clopen in $\cT_B$ and
|
||||
$\cB(X,\cT) = \cB(X, \cT_B)$.
|
||||
|
||||
Therefore $(\cB, \cT_B\defon{B})$ is Polish.
|
||||
We know that there is an embedding
|
||||
$f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$.
|
||||
|
||||
Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$.
|
||||
This is still continuous as $\cT \subseteq \cT_B$.
|
||||
Since $2^{\omega}$ is compact, $f$ is an embedding.
|
||||
%\todo{Think about this}
|
||||
\end{proof}
|
||||
|
||||
\begin{refproof}{thm:clopenize}
|
||||
We show that
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\
|
||||
&& (X, \cT_B) \text{ is Polish},\\
|
||||
&& \cB(X, \cT) = \cB(X, \cT_B)\\
|
||||
&& B \text{ is clopen in $\cT_B$}\\
|
||||
\}
|
||||
\end{IEEEeqnarray*}
|
||||
is equal to the set of Borel sets.
|
||||
|
||||
The proof rests on two lemmata:
|
||||
\begin{lemma}
|
||||
\label{thm:clopenize:l1}
|
||||
Let $(X,\cT)$ be a Polish space.
|
||||
Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$)
|
||||
there is $\cT_F \supseteq \cT$
|
||||
such that $\cT_F$ is Polish,
|
||||
$\cB(\cT) = \cB(\cT_F)$
|
||||
and $F$ is clopen in $\cT_F$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
|
||||
Both are Polish spaces.
|
||||
Take the coproduct\footnote{topological sum} $F \oplus (X \setminus F)$
|
||||
of these spaces.
|
||||
This space is Polish,
|
||||
and the topology is generated by $\cT \cup \{F\}$,
|
||||
hence we do not get any new Borel sets.
|
||||
\end{proof}
|
||||
So all closed sets are in $A$.
|
||||
Furthermore $A$ is closed under complements,
|
||||
since complements of clopen sets are clopen.
|
||||
|
||||
\begin{lemma}
|
||||
\label{thm:clopenize:l2}
|
||||
Let $(X, \cT)$ be Polish.
|
||||
Let $\{\cT_n\}_{n < \omega}$
|
||||
be Polish topologies
|
||||
such that $\cT_n \supseteq \cT$
|
||||
and $\cB(\cT_n) = \cB(\cT)$.
|
||||
Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
|
||||
is still Polish
|
||||
and $\cB(\cT_\infty) = \cB(T)$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
We have that $\cT_\infty$ is the smallest
|
||||
topology containing all $\cT_n$.
|
||||
To get $\cT_\infty$
|
||||
consider
|
||||
\[
|
||||
\cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}.
|
||||
\]
|
||||
Then
|
||||
\[
|
||||
\cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}.
|
||||
\]
|
||||
(It suffices to take countable unions,
|
||||
since we may assume that the $A_1, \ldots, A_n$ in the
|
||||
definition of $\cF$ belong to
|
||||
a countable basis of the respective $\cT_n$).
|
||||
|
||||
\todo{This proof will be finished in the next lecture}
|
||||
\end{proof}
|
||||
|
||||
We need to show that $A$ is closed under countable unions.
|
||||
By \yaref{thm:clopenize:l2} there exists a topology
|
||||
$\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$
|
||||
and $\cB(\cT_\infty) = \cB(\cT)$.
|
||||
Applying \yaref{thm:clopenize:l1}
|
||||
yields a topology $\cT_\infty'$ such that
|
||||
$(X, \cT_\infty')$ is Polish,
|
||||
$\cB(\cT_\infty') = \cB(\cT)$
|
||||
and $A $ is clopen in $\cT_{\infty}'$.
|
||||
\end{refproof}
|
||||
|
||||
|
||||
143
inputs/lecture_08.tex
Normal file
143
inputs/lecture_08.tex
Normal file
@ -0,0 +1,143 @@
|
||||
\lecture{08}{2023-11-10}{}
|
||||
|
||||
\todo{put this lemma in the right place}
|
||||
\begin{lemma}[Lemma 2]
|
||||
Let $(X, \cT)$ be a Polish space.
|
||||
Let $\cT_n \supseteq \cT$ be Polish
|
||||
with $\cB(X, \cT_n) = \cB(X, \cT)$.
|
||||
Let $\cT_\infty$ be the topology generated
|
||||
by $\bigcup_n \cT_n$.
|
||||
Then $(X, \cT_\infty)$ is Polish
|
||||
and $\cB(X, \cT_\infty) = \cB(X, \cT)$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Let $Y = \prod_{n \in \N} (X, \cT_n)$.
|
||||
Then $Y$ is Polish.
|
||||
Let $\delta\colon (X, \cT_\infty) \to Y$
|
||||
defined by $\delta(x) = (x,x,x,\ldots)$.
|
||||
\begin{claim}
|
||||
$\delta$ is a homeomorphism.
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
Clearly $\delta$ is a bijection.
|
||||
We need to show that it is continuous and open.
|
||||
|
||||
Let $U \in \cT_i$.
|
||||
Then
|
||||
\[
|
||||
\delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
|
||||
\]
|
||||
hence $\delta$ is continuous.
|
||||
Let $U \in \cT_\infty$.
|
||||
Then $U$ is the union of sets of the form
|
||||
\[
|
||||
V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
|
||||
\]
|
||||
for some $n_1 < n_2 < \ldots < n_u$
|
||||
and $U_{n_i} \in \cT_i$.
|
||||
|
||||
Thus is suffices to consider sets of this form.
|
||||
We have that
|
||||
\[
|
||||
\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.
|
||||
\]
|
||||
\end{subproof}
|
||||
|
||||
This will finish the proof since
|
||||
\[
|
||||
D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
|
||||
\]
|
||||
Why? Let $(x_n) \in Y \setminus D$.
|
||||
Then there are $i < j$ such that $x_i \neq x_j$.
|
||||
Take disjoint open $x_i \in U$, $x_j \in V$.
|
||||
Then
|
||||
\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]
|
||||
is open in $Y\setminus D$.
|
||||
Hence $Y \setminus D$ is open, thus $D$ is closed.
|
||||
It follows that $D$ is Polish.
|
||||
\end{proof}
|
||||
|
||||
\subsection{Parametrizations}
|
||||
\todo{choose better title}
|
||||
|
||||
|
||||
Let $\Gamma$ denote a collection of sets in some space.
|
||||
For us $\Gamma$ will be one of $\Sigma^0_\xi(X), \Pi^0_\xi(X), \Delta^0_\xi(X), \cB(X)$,
|
||||
where $X$ is a metrizable, usually second countable space.
|
||||
|
||||
\begin{definition}
|
||||
We say that $\cU \subseteq Y \times X$
|
||||
is \vocab{$Y$-universal} for $\Gamma(X)$ /
|
||||
$\cU$ \vocab{parametrizes} $\Gamma(X)$
|
||||
iff:
|
||||
\begin{itemize}
|
||||
\item $\cU \in \Gamma$,
|
||||
\item $\{U_y : y \in Y\} = \Gamma(X)$.
|
||||
\end{itemize}
|
||||
\end{definition}
|
||||
|
||||
\begin{example}
|
||||
Let $X = \omega^\omega$, $Y = 2^{\omega}$
|
||||
and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$.
|
||||
We will show that there is a $2^{\omega}$-universal
|
||||
set for $\Gamma$.
|
||||
\end{example}
|
||||
|
||||
\begin{theorem}
|
||||
Let $X$ be a separable, metrizable space.
|
||||
Then for every $\xi \ge 1$,
|
||||
there is a $2^{\omega}$-universal
|
||||
set for $\Sigma^0_\xi(X)$ and
|
||||
similarly for $\Pi^0_\xi(X)$.
|
||||
\end{theorem}
|
||||
\begin{proof}
|
||||
Note that if $\cU$ is $2^{\omega}$ universal for
|
||||
$\Sigma^0_\xi(X)$, then $(2^{\omega} \times X) \setminus \cU$
|
||||
is $2^{\omega}$-universal for $\Pi^0_\xi(X)$.
|
||||
Thus it suffices to consider $\Sigma^0_\xi(X)$.
|
||||
|
||||
First let $\xi = 1$.
|
||||
We construct $\cU \overset{\text{open}}{\subseteq} 2^{\omega} \times X$
|
||||
such that
|
||||
\[
|
||||
\{U_y : y \in 2^\omega\} = \Sigma^0_1(X).
|
||||
\]
|
||||
|
||||
Let $(V_n)$ be a basis of open sets of $X$.
|
||||
For all $y \in 2^\omega$ and $x \in X$
|
||||
put $(y,x) \in \cU$ iff
|
||||
$x \in \bigcup \{V_n : y_n = 1\}$.
|
||||
$\cU$ is open.
|
||||
Let $V = \bigcup \{V_n : V_n \subseteq V\}$.
|
||||
Pick $y \in 2^\omega$
|
||||
and let $y_n = 1$ iff $V_n \subseteq V$.
|
||||
Then $\cU_y = V$.
|
||||
|
||||
|
||||
Now suppose that there exists a
|
||||
$2^{\omega}$-universal set for $\Sigma^0_{\eta}(X)$
|
||||
for all $\eta < \xi$.
|
||||
Fix $\xi_0 \le \xi_1 \le \ldots < \xi$
|
||||
such that $\xi_n \to \xi$ if $\xi$ is a limit,
|
||||
or $\xi_n = \xi'$ if $\xi = \xi' +1$ is a successor.
|
||||
|
||||
Recall that $\eta_1 \le \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$.
|
||||
|
||||
Note that if $A = \bigcup_n A_n$, with $A_n \in \Pi^0_{\eta_n}(X)$
|
||||
some $\eta_n < \xi$,
|
||||
we also have
|
||||
$A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$.
|
||||
|
||||
We construct a $(2^\omega)^\omega \cong 2^\omega$-universal set
|
||||
for $\Sigma^0_\xi(X)$.
|
||||
For $(y_n) \in (2^\omega)^\omega$
|
||||
and $x \in X$
|
||||
we set $((y_n), x) \in \cU$
|
||||
iff $\exists n.~(y_n, x) \in U_{\xi_n}$,
|
||||
i.e.~iff $\exists n.~x \in (U_{\xi_n})_{y_n}$.
|
||||
|
||||
Let $A \in \Sigma^0_\xi(X)$.
|
||||
Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
|
||||
% TODO
|
||||
Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$.
|
||||
\end{proof}
|
||||
@ -134,5 +134,6 @@
|
||||
\DeclareMathOperator{\hght}{height}
|
||||
\DeclareMathOperator{\symdif}{\triangle}
|
||||
\DeclareSimpleMathOperator{proj}
|
||||
\newcommand{\fc}{\mathfrak{c}}
|
||||
|
||||
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
|
||||
|
||||
@ -30,6 +30,8 @@
|
||||
\input{inputs/lecture_04}
|
||||
\input{inputs/lecture_05}
|
||||
\input{inputs/lecture_06}
|
||||
\input{inputs/lecture_07}
|
||||
\input{inputs/lecture_08}
|
||||
|
||||
|
||||
|
||||
|
||||
Loading…
Reference in New Issue
Block a user