# Rudin 2.14

1. Jun 29, 2015

### Rasalhague

1. The problem statement, all variables and given/known data

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.)

2. Relevant equations

Every separable metric space has a countable base.

3. 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?

2. Jun 29, 2015

### 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.

3. Jun 29, 2015

### micromass

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

4. Jun 29, 2015

### WWGD

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

5. Jun 29, 2015

### micromass

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

6. Jun 30, 2015

### Rasalhague

Got it, thanks!