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

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