## Cantor-Bendixson decompositions. Kechris, Exercise (6.6)

Kechris, p. 32, Exercise (6.6) (A)
Let $X$ be a Polish space. Decompose $X$ uniquely into $P \dot{\cup} C$, where $P$ is perfect, $C$ is countable open. This decomposition is possible by Cantor-Bendixson Theorem.

Exercise  (6.6): $P$ is the largest perfect subset of $X$.

Proof.
Suppose $Q \subset X$ is perfect (so closed and perfect in its relative topology).  From the proof of Theorem (6.4), and in the notation defined there, we have $Q^* = Q$, hence $Q \subset P$.  Now suppose $Q \supset P$, and further that $Q \cap C \neq \emptyset$; notice that $Q \cap C$, being a $G_\delta$ set, is also a Polish space.

Claim: $Q \cap C$ is a perfect space.
Let $x \in Q \cap C$.  Let $V$ be an open nbhd of $x \in Q \cap C$.  That is, there exists $V'$ open in $X$ such that $V = V' \cap Q \cap C$.  Then $V$ is also open in $Q$, since $C \cap V'$ is open in $X$.  Since $Q$ is perfect, there exists $y \neq x \in V$. Therefore, $Q \cap C$ is perfect and nonempty, hence uncountable by Theorem (6.2).

But this contradicts that $C$ is countable.  Hence, $Q \cap C = \emptyset$, and thus $Q = P$. $\blacksquare$