Proving that Every Closed Set in Separable Metric Space is Union of Perfect and Countable Set

[Rasalhague]

Homework Statement

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. (Rudin: Principles of Mathematical Analysis, 2nd ed.)

Homework Equations

Every separable metric space has a countable base.

The Attempt at a Solution

Let F be closed. Using the above fact, I've shown that the isolated points of F are at most countable, likewise their closure. I'm trying to construct a perfect set by removing non-limit points of F' points from F', the set of limit points of F, but it's not quite falling into place yet. Is this a good direction to go?

 

[WWGD]

Careful, that is a countable _local_ base, not a countable global base, i.e., every metric space is 1st-countable, but not necessarily 2nd-countable.
EDIT: Sorry, I did not read carefully: the hypothesis of separable implies 2nd countable.

 

[micromass]

Careful, that is a countable _local_ base, not a countable global base, i.e., every metric space is 1st-countable, but not necessarily 2nd-countable.

It's a countable global base since the space is separable.

 

[WWGD]

Yes, thanks, I just corrected it as you were writing; I did not read carefully-enough.

 

[micromass]

Hint: consider the points $x\in F$ for which every neighborhood of $x$ meets $F$ in uncountably many points.

 

[Rasalhague]

Got it, thanks!