# Complete separable metric space X represented represented as union of closed sets

I have a problem concerning a statement I found in volume 2 of the classic reference book on measure theory by Bogachev. More precisely, I have a problem concerning theorem 6.1.13.

I the proof the author states that a nonempty complete separable metric space $X$ can be represented in the form $X = 􏰳\bigcup_{j=1}^\infty E(j)$, where the sets $E(j)$ are closed (not necessarily disjoint) of diameter less than $2^{−3}$. What is the reason?

My guess: The reason is that $X$ is separable, hence it has a countable base, and what works for open sets, complementary works for closed as well. Hence, we have a countable base that corresponds to those sets.

Is my intuition correct?

Thank you for your time.
Any feedback is most welcome.

## 1 Answer

You can't really carry statements about open sets over to statements about closed sets. On the other hand, the countable base is what makes this work. If $D$ is a countable dense set you have $$X = \bigcup_{d \in D} B[d,2^{-4}]$$ where $B[d,2^{-4}]$ is the closed ball of radius $2^{-4}$ centered at $d$.

• Thanks a lot. May I ask you to expand a bit your first statement concerning open and closed sets, and why here it does not work? I ask you this, because I think that if you just make your balls open, still it works (of course by starting all over with open $E(j)$), and precisely for the same reason, i.e. the countable dense set and the corresponding countable base. – Kolmin Apr 29 '15 at 18:21
• In this context, yes, you can use either open balls or closed balls. However, your OP states "what works for open sets, complementary works for closed as well" which is in general false. – Umberto P. Apr 29 '15 at 19:25
• Yes, I know. True enough, in this context it is actually the same, because we can define topological spaces as built upon either open spaces, or closed spaces, and in this case we start from closed sets, thus the basis is "characterized" by closed sets. Correct me if I am wrong, but in this kind of situations, talking about closed or open sets - as a matter of fact - does not make much difference. Of course there is a huge difference, as you probably pointed out in your original answer, if we ask ourselves if $(0,1)$ or $[0,1]$ are compact! (This is an example of what you meant, right?) – Kolmin Apr 29 '15 at 20:55