# If P is a closed perfect subset of a separable complete metric space M, then each point of P is a condensation point of P.

Pugh's Real Mathematical Analysis, first edition, p. 131, problem 108:

Let $P$ be a closed perfect subset of a separable complete metric space $M$. Prove that each point of $P$ is a condensation point of $P$.

I think I can prove this, but I never use the hypothesis that $M$ is separable, so I wonder if I've done something wrong. Here is my attempt:

Proposition. If $P$ is a closed perfect subset of a complete metric space $M$, then each point of $P$ is a condensation point of $P$.

Proof. Suppose not, so we have some point $p \in P$ which is not a condensation point of $P$, and thus has a countable neighborhood $P_r p$ of radius $r > 0$. Since there are only countably many points in this neighborhood, there must be some real radius $s \in (0, r]$ such that no points in $P$ are distance $s$ from $p$, and thus $P_s p$ has empty boundary in $P$ and is therefore clopen in $P$.

An open subset of a perfect set is perfect, so $P_s p \subseteq P$ is perfect. A closed subset of a complete set is complete, so $P_s p \subseteq P \subseteq M$ is complete. $P_s p$ is non-empty (since it contains $p$), perfect, and complete, so by Theorem 58 it is uncountable. However, $P_s p$ is a subset of $P_r p$, so this implies that $P_r p$ is uncountable, a contradiction. $\Box$

Have I made some mistake? Thanks very much.

• What is a "condensation point"? Feb 28, 2016 at 15:48
• A point p of S is a condensation point if every neighborhood of p in S contains uncountably many points. Feb 28, 2016 at 20:48
• Your proof would be self-contained except for "Theorem 58"; other than that it looks fine. Feb 28, 2016 at 22:10
• Thanks! The proof of theorem 58 uses diagonalization to produce a Cauchy sequence converging to a point not on the proposed enumeration. Feb 28, 2016 at 22:21
• Seems pretty clear then. I cannot see anything wrong. Feb 29, 2016 at 13:58