Josia Pietsch
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\lecture{07}{20231107}{}




\begin{proposition}


Let $X$ be second countable.


Then $\cB(X) \le \fc$.


% $\fc := 2^{\aleph_0}$


\end{proposition}


\begin{proof}


\gist{%


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.


\]


}{Use strong induction. $\Sigma^0_1(X) \le \fc$, since $X$ is second countable.


\[\Sigma^0_\xi(X) \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}\]


}


\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.


\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.


\item $\Delta^0_\xi(X)$ is closed under finite unions and


finite intersections.


\end{itemize}




\end{enumerate}


\end{proposition}


\gist{%


\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%


\gist{ (countable unions yield all open sets, but there are open


sets that are not clopen)}{}.


\end{example}




\subsection{Turning Borel Sets into Clopens}




\begin{theorem}%


\gist{%


\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''


unfortunately seems to be nonstandard vocabulary.


Our tutor repeatedly advised against using it in the final exam.


Contrary to popular belief


the very same tutor was \textit{not} the one first to introduce it,


as it would certainly be spelled ``to clopenise'' if that were the case.


}%


}{}%


\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}


\gist{%


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.


}{%


Clopenize $B$.


We can embed $2^{ \omega}$ into Polish spaces.


Clopenization makes the topology finer,


so this is still continuous wrt.~the original topology.


$2^{\omega}$ is compact, so this is an embedding.


}


\end{proof}




\begin{refproof}{thm:clopenize}


\gist{%


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.


}{%


Let $A$ be the set of clopenizable sets.


We show that $A = \cB(X)$.


}


\gist{The proof rests on two lemmata:}{}


\begin{lemma}


\label{thm:clopenize:l1}


\gist{%


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 $\cT_F$.


}{%


Closed sets can be clopenized.


}


\end{lemma}


\begin{proof}


\gist{%


Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.


Both are Polish spaces.


Take the coproduct%


\footnote{In the lecture, this was called the \vocab{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.


}{Consider $(F, \cT\defon{F}) \oplus (X \setminus F, \cT\defon{X \setminus F})$.}


\end{proof}


\gist{%


So all closed sets are in $A$.


Furthermore $A$ is closed under complements,


since complements of clopen sets are clopen.


}{So $\Sigma^0_1(X), \Pi^0_1(X) \subseteq A$.}




\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 Polish


and $\cB(\cT_\infty) = \cB(T)$.


\end{lemma}


\begin{refproof}{thm:clopenize:l2}


\gist{%


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 $\cT_n$).


}{}




% Proof was finished in lecture 8


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}


\gist{%


\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_{n_u}


\]


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}


}{}




\begin{claim}


$D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y.$


\end{claim}


\gist{%


\begin{subproof}


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.


\end{subproof}


It follows that $D$ is Polish.


}{}


\end{refproof}




\gist{%


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}
