Josia Pietsch 458dd9ab1f
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\subsection{Sheet 3}
\nr 1
Let $A \neq \emptyset$ be discrete.
For $D \subseteq A^{\omega}$,
T_D \coloneqq \{x\defon{n} \in A^{<\omega} | x \in D, n \in \N\}.
\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 $A^{\omega}$.
$[T_D] = D$ for any closed subset $D \subseteq A^{\omega}$.
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 $x \in D$.
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 $X_i \subseteq A^{n_i}$, $n_i < \omega$.
$U = \bigcup_{s \in S'} N_s$
$S' \coloneqq \bigcup_{i \in I} X_i$.
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
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:
% https://alanmath.wordpress.com/2011/06/16/on-trees-compactness-and-finite-splitting/
\nr 2
\nr 3
\nr 4
For $A \subseteq X$ let $A'$ denote the set of
accumulation points of $A$.
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 $U$.
Let $P$ be the set of condensation points of $X$
and $U \coloneqq X \setminus P$.
$U$ is open and countable.
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.
$P$ is perfect.
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 $x \in P$.
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 $U_2 = U$.
Any Polish space is either countable or has cardinality equal to $\fc$.
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}.