## Characterization of separability

Prove that a metric space is separable if and only if there does not exist an uncountable set such that for every .

This is a useful property which makes trivial the proof of the fact that any subset of a separable space is separable.

*Proof:* One implication is trivial. If is separable, then there cannot exist an uncountable set with pairwise “big” distances. The argument would be that if is separable, then in each ball there should be an element of the countable dense set, and this is impossible, since this is an uncountable family of disjoint balls.

Suppose now that there does not exist an uncountable set with big enough pairwise distances. That means that whenever we find a set such that for every we have that is at most countable.

Pick now and denote

For every we know that os not empty (pick for example a two point set, and we will prove using Zorn’s lemma that each has maximal elements. Indeed, order each by inclusion and consider a totally ordered subfamily . Pick now . Obviously is an upper bound for every , and moreover, , since if then there exist such that and by total order assumption we have, for example , which means that . Since each chain has an upper bound, it follows by Zorn’s lemma that the set admits maximal elements. (a maximal element has the property that whenever we necessarily have ; note that maximal elements are not necessarily unique.)

For each rational, consider a maximal element of . Then is a countable subset of which I claim is dense. To prove this pick . Since is maximal, there exists at least one such that . We do this for every positive rational and we find a sequence with the property that . Taking we see that . By the arbitrairiness of we can say that is dense in and therefore is separable.