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.
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 $d$.