Josia Pietsch
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278 lines
10 KiB
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278 lines
10 KiB
TeX
\lecture{07}{2023-11-07}{}
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\begin{proposition}
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Let $X$ be second countable.
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Then $|\cB(X)| \le \fc$.
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% $\fc := 2^{\aleph_0}$
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\end{proposition}
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\begin{proof}
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\gist{%
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We use strong induction on $\xi < \omega_1$.
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We have $\Sigma^0_1(X) \le \fc$
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(for every element of the basis, we can decide
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whether to use it in the union or not).
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Suppose that $\forall \xi' < \xi.~|\Sigma^0_{\xi'}(X)| \le \fc$.
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Then $|\Pi^0_{\xi'}(X)| \le \fc$.
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We have that
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\[
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\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
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\]
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Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.
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We have
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\[
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\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).
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\]
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Hence
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\[
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|\cB(X)| \le \omega_1 \cdot \fc = \fc.
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\]
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}{Use strong induction. $|\Sigma^0_1(X)| \le \fc$, since $X$ is second countable.
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\[|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}\]
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}
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\end{proof}
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\begin{proposition}[Closure properties]
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Suppose that $X$ is metrizable.
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Let $1 \le \xi < \omega_1$.
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Then
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\begin{enumerate}[(a)]
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\item \begin{itemize}
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\item $\Sigma^0_\xi(X)$ is closed under countable unions.
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\item $\Pi^0_\xi(X)$ is closed under countable intersections.
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\item $\Delta^0_\xi(X)$ is closed under complements.
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\end{itemize}
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\item \begin{itemize}
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\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
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\item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
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\item $\Delta^0_\xi(X)$ is closed under finite unions and
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finite intersections.
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\end{itemize}
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\end{enumerate}
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\end{proposition}
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\gist{%
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\begin{proof}
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\begin{enumerate}[(a)]
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\item This follows directly from the definition.
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Note that a countable intersection can be written
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as a complement of the countable union of complements:
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\[
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\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
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\]
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\item If suffices to check this for $\Sigma^0_{\xi}(X)$.
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Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
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and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
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Then
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\[
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A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)
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\]
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and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
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\end{enumerate}
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\end{proof}
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}{}
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\begin{example}
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Consider the cantor space $2^{\omega}$.
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We have that $\Delta^0_1(2^{\omega})$
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is not closed under countable unions%
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\gist{ (countable unions yield all open sets, but there are open
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sets that are not clopen)}{}.
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\end{example}
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\subsection{Turning Borel Sets into Clopens}
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\begin{theorem}%
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\gist{%
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\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
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unfortunately seems to be non-standard vocabulary.
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Our tutor repeatedly advised against using it in the final exam.
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Contrary to popular belief
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the very same tutor was \textit{not} the one first to introduce it,
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as it would certainly be spelled ``to clopenise'' if that were the case.
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}%
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}{}%
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\label{thm:clopenize}
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Let $(X, \cT)$ be a Polish space.
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For any Borel set $A \subseteq X$,
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there is a finer Polish topology,%
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\footnote{i.e.~$\cT_A \supseteq \cT$ and $(X, \cT_A)$ is Polish}
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such that
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\begin{itemize}
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\item $A$ is clopen in $\cT_A$,
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\item the Borel sets do not change,
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i.e.~$\cB(X, \cT) = \cB(X, \cT_A)$.
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\end{itemize}
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\end{theorem}
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\begin{corollary}[Perfect set property]
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Let $(X, \cT)$ be Polish,
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and let $B \subseteq X$ be Borel and uncountable.
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Then there is an embedding
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of the cantor space $2^{\omega}$
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into $B$.
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\end{corollary}
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\begin{proof}
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\gist{%
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Pick $\cT_B \supset \cT$
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such that $(X, \cT_B)$ is Polish,
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$B$ is clopen in $\cT_B$ and
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$\cB(X,\cT) = \cB(X, \cT_B)$.
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Therefore $(\cB, \cT_B\defon{B})$ is Polish.
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We know that there is an embedding
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$f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$.
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Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$.
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This is still continuous as $\cT \subseteq \cT_B$.
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Since $2^{\omega}$ is compact, $f$ is an embedding.
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}{%
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Clopenize $B$.
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We can embed $2^{ \omega}$ into Polish spaces.
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Clopenization makes the topology finer,
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so this is still continuous wrt.~the original topology.
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$2^{\omega}$ is compact, so this is an embedding.
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}
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\end{proof}
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\begin{refproof}{thm:clopenize}
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\gist{%
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We show that
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\begin{IEEEeqnarray*}{rCl}
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A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\
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&& (X, \cT_B) \text{ is Polish},\\
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&& \cB(X, \cT) = \cB(X, \cT_B)\\
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&& B \text{ is clopen in $\cT_B$}\\
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\}
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\end{IEEEeqnarray*}
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is equal to the set of Borel sets.
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}{%
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Let $A$ be the set of clopenizable sets.
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We show that $A = \cB(X)$.
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}
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\gist{The proof rests on two lemmata:}{}
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\begin{lemma}
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\label{thm:clopenize:l1}
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\gist{%
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Let $(X,\cT)$ be a Polish space.
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Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$)
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there is $\cT_F \supseteq \cT$
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such that $\cT_F$ is Polish,
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$\cB(\cT) = \cB(\cT_F)$
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and $F$ is clopen in $\cT_F$.
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}{%
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Closed sets can be clopenized.
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}
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\end{lemma}
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\begin{proof}
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\gist{%
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Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
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Both are Polish spaces.
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Take the coproduct%
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\footnote{In the lecture, this was called the \vocab{topological sum}.}
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$F \oplus (X \setminus F)$ of these spaces.
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This space is Polish,
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and the topology is generated by $\cT \cup \{F\}$,
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hence we do not get any new Borel sets.
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}{Consider $(F, \cT\defon{F}) \oplus (X \setminus F, \cT\defon{X \setminus F})$.}
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\end{proof}
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\gist{%
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So all closed sets are in $A$.
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Furthermore $A$ is closed under complements,
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since complements of clopen sets are clopen.
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}{So $\Sigma^0_1(X), \Pi^0_1(X) \subseteq A$.}
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\begin{lemma}
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\label{thm:clopenize:l2}
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Let $(X, \cT)$ be Polish.
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Let $\{\cT_n\}_{n < \omega}$
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be Polish topologies
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such that $\cT_n \supseteq \cT$
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and $\cB(\cT_n) = \cB(\cT)$.
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Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
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is Polish
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and $\cB(\cT_\infty) = \cB(T)$.
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\end{lemma}
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\begin{refproof}{thm:clopenize:l2}
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\gist{%
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We have that $\cT_\infty$ is the smallest
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topology containing all $\cT_n$.
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To get $\cT_\infty$ consider
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\[
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\cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}.
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\]
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Then
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\[
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\cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}.
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\]
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(It suffices to take countable unions,
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since we may assume that the $A_1, \ldots, A_n$ in the
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definition of $\cF$ belong to
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a countable basis of the respective $\cT_n$).
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}{}
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% Proof was finished in lecture 8
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Let $Y = \prod_{n \in \N} (X, \cT_n)$.
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Then $Y$ is Polish.
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Let $\delta\colon (X, \cT_\infty) \to Y$
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defined by $\delta(x) = (x,x,x,\ldots)$.
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\begin{claim}
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$\delta$ is a homeomorphism.
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\end{claim}
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\gist{%
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\begin{subproof}
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Clearly $\delta$ is a bijection.
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We need to show that it is continuous and open.
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Let $U \in \cT_i$.
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Then
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\[
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\delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
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\]
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hence $\delta$ is continuous.
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Let $U \in \cT_\infty$.
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Then $U$ is the union of sets of the form
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\[
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V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{n_u}
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\]
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for some $n_1 < n_2 < \ldots < n_u$
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and $U_{n_i} \in \cT_i$.
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Thus is suffices to consider sets of this form.
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We have that
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\[
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\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.
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\]
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\end{subproof}
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}{}
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\begin{claim}
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$D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y.$
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\end{claim}
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\gist{%
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\begin{subproof}
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Let $(x_n) \in Y \setminus D$.
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Then there are $i < j$ such that $x_i \neq x_j$.
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Take disjoint open $x_i \in U$, $x_j \in V$.
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Then
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\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]
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is open in $Y\setminus D$.
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Hence $Y \setminus D$ is open, thus $D$ is closed.
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\end{subproof}
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It follows that $D$ is Polish.
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}{}
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\end{refproof}
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\gist{%
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We need to show that $A$ is closed under countable unions.
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By \yaref{thm:clopenize:l2} there exists a topology
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$\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$
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and $\cB(\cT_\infty) = \cB(\cT)$.
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Applying \yaref{thm:clopenize:l1}
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yields a topology $\cT_\infty'$ such that
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$(X, \cT_\infty')$ is Polish,
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$\cB(\cT_\infty') = \cB(\cT)$
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and $A $ is clopen in $\cT_{\infty}'$.
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}{}
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\end{refproof}
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