Josia Pietsch
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\subsection{Sheet 3}




\tutorial{04}{}{}




\nr 1




Let $A \neq \emptyset$ be discrete.


For $D \subseteq A^{\omega}$,


let


\[


T_D \coloneqq \{x\defon{n} \in A^{<\omega}  x \in D, n \in \N\}.


\]


\begin{enumerate}[(a)]


\item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:




Clearly $T_D$ is a tree.


Let $x \in T_D$.


Then there exists $d \in D$ such that $x = d\defon{n}$.


Hence $x \subseteq d\defon{n+1} \in T_D$.


Thus $x$ is not a leaf, i.e.~$T_D$ is pruned.


\item For any $T \subseteq A^{<\omega}$, $[T]$ is a closed subset


of $A^{\omega}$:




Let $a \in A^{\omega} \setminus [T]$.


Then there exists some $n$ such that $a\defon{n} \not\in T$.


Hence $\{a_0\} \times \ldots \times \{a_{n+1}\} \times A^{\omega}$


is an open neighbourhood of $a$ disjoint from $[T]$.




\item $T \mapsto [T]$ is a bijection between the


pruned trees on $A$ and the closed subsets of $A^{\omega}$.




\begin{claim}


$[T_D] = D$ for any closed subset $D \subseteq A^{\omega}$.


\end{claim}


\begin{subproof}


Clearly $D \subseteq [T_D]$.


Let $x \in [T_D]$.


Then for every $n < \omega$,


there exists some $d_n \in D$ such that


$d_n\defon{n} = x\defon{n}$.


Clearly the $d_n$ converge to $x$.


Since $D$ is closed, we get $x \in D$.


\end{subproof}




This shows that $T \mapsto [T]$ is surjective.




Now let $T \neq T'$ be pruned trees.


Then there exists $x \in T \mathop{\triangle} T'$,


wlog.~$x \in T \setminus T'$.


Since $T$ is pruned


by applying the axiom of countable choice


we get an infinite branch $x' \in [T] \setminus [T']$.


Hence the map is injective.




\item Let $N_s \coloneqq \{x \in A^{\omega}  s \subseteq x\}$.


Show that every open $U \subseteq A^{\omega}$


can be written as $U = \bigcup_{s \in S} N_s$


for some set of pairwise incompatible $S \subseteq A^{<\omega}$.






Let $U$ be open.


Then $U$ has the form


\[


U = \bigcup_{i \in I} X_i \times A^{\omega}


\]


for some $X_i \subseteq A^{n_i}$, $n_i < \omega$.


Clearly


$U = \bigcup_{s \in S'} N_s$


for


$S' \coloneqq \bigcup_{i \in I} X_i$.


Define


\[


S \coloneqq \{s \in S'  \lnot\exists t \in S'.~t\subseteq s \land t < s\}.


\]


Then the elements of $S$ are pairwise incompatible


and $U = \bigcup_{s \in S} N_s$.




\item Let $T \subseteq A^{<\omega}$ be an infinite tree which is finitely


splitting.


Then $[T]$ is nonempty:




Let us recursively construct a sequence of compatible $s_n \in T$


with $s_n = n$


such that $\{s_n\} \times A^{<\omega} \cap T$ is infinite.


Let $s_0$ be the empty sequence;


by assumption $T$ is infinite.


Suppose that $s_n$ has been chosen.


Since $T$ is finitely splitting, there are only finitely


many $a \in A$ with $s_n\concat a \in T$.


Since $ \{s_n\} \times A^{<\omega} \cap T$


is infinite,


there must exist at least on $a \in A$


such that $\{s_n\concat a\} \times A^{<\omega} \cap T$


is infinite.


Define $s_{n+1} \coloneqq s_n \concat a$.




Then the union of the $s_n$ is an infinite branch of $T$,


i.e.~$[T]$ is nonempty.






\item Then $[T]$ is compact:


\todo{TODO}




% https://alanmath.wordpress.com/2011/06/16/ontreescompactnessandfinitesplitting/


\end{enumerate}




\nr 2


\todo{handwritten}


\nr 3


\todo{handwritten}




\nr 4




\begin{notation}


For $A \subseteq X$ let $A'$ denote the set of


accumulation points of $A$.


\end{notation}




\begin{theorem}


Let $X$ be a Polish space.


Then there exists a unique partition $X = P \sqcup U$


of $X$ into a perfect closed subset $P$ and a countable open subset $U$.


\end{theorem}


\begin{proof}




Let $P$ be the set of condensation points of $X$


and $U \coloneqq X \setminus P$.




\begin{claim}


$U$ is open and countable.


\end{claim}


\begin{subproof}


Let $S$ be a countable dense subset.


For each $x \in U$,


there is an $\epsilon_x > 0$, $s_x \in S$ such that $x \in B_{\epsilon_x}(s_x)$


is at most countable.


Clearly $B_{\epsilon_x}(s_x) \subseteq U$,


as for every $y \in B_{\epsilon_x}(s_x)$,


$B_{\epsilon_x}(s_x)$ witnesses that $y \not\in P$.


Thus $U = \bigcup_{x \in U} B_{\epsilon_x}(s_x)$


is open.


Wlog.~$\epsilon_x \in \Q$ for all $x$.


Then the RHS is the union of at most


countably many countable sets, as $S \times \Q$


is countable.


\end{subproof}




\begin{claim}


$P$ is perfect.


\end{claim}


\begin{subproof}


Let $x \in P'$ and $x \in U$ an open neighbourhood.


Then there exists $y \in P \cap U$.


In particular, $U$ is an open neighbourhood of $ y$,


hence $U$ is uncountable.


It follows that $x \in P$.




On the other hand let $x \in P$


and let $U$ be an open neighbourhood.


We need to show that $U \cap P \setminus \{x\}$


is not empty.


Suppose that for all $y \in U \cap P \setminus \{x\}$,


there is an open neighbourhood $U_y$


such that $U_y$ is at most countable.


Wlog.~$U_y = B_{\epsilon_y}(s_y)$ for some $s_y \in S$, $\epsilon_y > 0$,


where $S$ is again a countable dense subset.


Wlog.~$\epsilon_y \in \Q$.


But then


\[


U = \{x\} \cup \bigcup_{y \in U} B_{\epsilon_y}(s_y)


\]


is at most countable as a countable union of countable sets,


contradiction $x \in P$.


\end{subproof}




\begin{claim}


Let $P,U$ be defined as above


and let $P_2 \subseteq X$, $U_2 \subseteq X$


be such that $P_2$ is perfect and closed,


$U_2$ is countable and open


and $X = P_2 \sqcup U_2$.


Then $P_2 = P$ and $U_2 = U$.


\end{claim}


\todo{TODO}


\end{proof}




\begin{corollary}\label{cor:polishcard}


Any Polish space is either countable or has cardinality equal to $\fc$.


\end{corollary}


\begin{subproof}


Let $X = P \sqcup U$


where $P$ is perfect and $U$ is countable.


If $P \neq \emptyset$, we have $P = \fc$


by \yaref{cor:perfectpolishcard}.


\end{subproof}
