# Is every set in a separable metric space the union of a perfect set and a set that is at most countable?

I'm reading Baby Rudin and exercise 28 of chapter 2 reads "Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable." I don't understand why the set must be closed in order for this to be true. My proof is as follows, and I don't make any reference to the closedness of the set:

Let $E$ be a set in a separable metric space. Let $B$ be a countable base of that metric space. Let $P$ be the set of condensation points of $E$. I claim that $P$ is perfect and $E-P$ is at most countable.

If $p$ is a limit point of $P$, then every neighborhood of $p$ includes at least one point $q$ of P. For a neighborhood $N_{p}$ of $p$, choose a neighborhood $N_{q} \subset N_{p}$ of $q$. Since $q$ is a condensation point of $E$, uncountably many points of $E$ are in $N_{q}$, and thus in $N_{p}$. So $p\in P$ and $P$ is closed.

If $p \in P$, then $p$ is a condensation point of $E$, so every neighborhood $N_{p}$ of $p$ must include uncountably many points of $E$. If $p$ is not a limit point of $P$, then there exists a neighborhood $N_{p}$ of $p$ such that $N_p$ includes no condensation points of $E$ except for $p$. This means that every point $q\in N_{p}-\{p\}$ must have a neighborhood $N_{q}$ that does not have uncountably many points of $E$. For each $b_{k}\in B \cap \mathscr{P}(N_{p})$, choose such a neighborhood $N_{q_{k}} \supset b_{k}$ of some point $q\in N_{p}-\{p\}$. Now $\{N_{q_{k}}\}$ covers $N_{p}$, and $\bigcup_{k=1}^{\infty}N_{q_{k}}\supset N_{p}$ has at most countably many points of $E$, for it is a union of countably many sets that have at most countably many points of $E$. This contradicts the assumption that $p\in P$. So $P$ is perfect.

Let $S=E-P$. There are no condensation points in $S$. So there exists an open cover $\{N_s\}$, where each $N_s$ is a neighborhood of $s \in S$ with at most countably many elements of $S$. For each $b_{k}\in B \cap \mathscr{P}(S)$, choose such a neighborhood $N_{s_{k}} \supset b_{k}$ of some point $s\in S$. Then we have a countable subcover $\{N_{p_{k}}\}$ of $S$. As $N_{p_{k}} \cap S$ is at most countable for all $k$, $S$ must be at most countable.

So what am I doing wrong? Or is Rudin's assumption of closedness extraneous?

• If $E$ is not closed, then in general $P \not \subset E$. – Daniel Fischer Aug 16 '13 at 19:09
• oh ok. That's what I did wrong. So $E$ must be closed to ensure that $P \subset E$. – Joshua Meyers Aug 16 '13 at 19:18
• Right. Now, of course you can regard $E$ as a metric space in its own right, and as a subspace of a separable metric space, it is separable too. And of course it's closed then. The set of condensation points is then $P_E = P \cap E$, which is perfect (as a subset of $E$, not in general as a subset of $X$, since perfect sets are by definition closed), and $E \setminus P_E$ is countable. – Daniel Fischer Aug 16 '13 at 19:29
• In your proof, why does p belong to $\cup N_{q_{k}}$? I think you just want a countable cover of $N_{p}-\{p\}$ – SihOASHoihd Jul 13 '17 at 18:02
• Also, I dont think it's true that every basis element in $N_{p}$ is contained in some $N_{q_{k}}$. – SihOASHoihd Jul 13 '17 at 18:17

## 2 Answers

As is customary, Rudin’s definition of perfect set includes the requirement that the set be closed. Your argument would be correct if that requirement were not included in the definition, though I think that it’s a little easier to make it the other way round: let $$E=\{x\in S:x\text{ has an open nbhd }B\text{ such that }B\cap S\text{ is countable}\}\;,$$ show that $E$ is countable, and show that every point of $S\setminus E$ is a condensation point of $S\setminus E$.

Following Brian's answer: If you do that for $E=\mathbb R\setminus\mathbb Q$, then $P=E$, which is not closed.