I am studying Kechris' proof of the Cantor-Bendixson theorem (from his "DST bible", I. 6.B (6.4)). The theorem says the following:
Let $(X,\tau)$ be a Polish (i.e. separable, completely metrisable) space. There exists a perfect set $P$ and a countable set $C$ such that $X = P \cup C$.
A point $x \in X$ is a limit point of $A \subset X$ if every open set $U \ni x$ contains some $y \in A$ such that $y \neq x$.
A condensation point satisfies that each of its open neighbourhoods is uncountable.
Kechris proposes to choose the set of condensation points in $X$ as the perfect set $P$. In order to show that $P$ has no isolated points, he takes some $x \in P$ and some open set $U$ containing $x$. Now, $U$ is uncountable since $x$ is a condensation point. So it suffices to show that $U$ contains some condensation point other than $x$. Now, Kechris simply mentions that
[since] $U$ is uncountable, […] it contains uncountably many condensation points
from which he easily deduces that $U \cap P$ is non-empty (even uncountable), which completes the proof of existence.
Why is this statement true?
Best Answer
$X$ is separable and metrizable, so it has a countable base $\mathscr{B}$. Let $V$ be the set of points of $X$ that are not condensation points. Each $x\in V$ has a countable open nbhd $B_x\in\mathscr{B}$, and it’s clear that $B_x\subseteq V$, so in fact $V=\bigcup\{B_x:x\in V\}$ and is countable. Take $P=X\setminus V$, the set of condensation points. Now fix $x\in P$, and let $U$ be an open nbhd of $x$; $U$ is uncountable, so $U\cap P=U\setminus V$ is also uncountable.