## Separable spaces

A Banach space is called separable if it contains a countable dense subset. Here are some interesting facts about separable spaces.

Prove that if is a subset of a separable metric space then is also separable.

If the dual of a normed vector space is separable, then the space itself is separable.

**Proof:** Let be a separable space and denote by a countable dense set of . We cannot just say that since is a subset of , then some part of is dense in , because it is possible that most of the elements of to be in . Therefore, we need another approach.

Then because is dense, we have that for any there exists such that . Denote , and prove that this is a countable dense set in . It is clearly a countable set, as the countable union of countable sets. Take and . There exists such that . So and for such that we have that

, which finishes the proof.

For the second part, let be a normed vector space and denote by its dual space. By Hahn-Banach’s Theorem, for every proper closed subspace there exists a linear functional which vanishes on , has norm and for some we have . Suppose now that is separable. Then its unit sphere is also separable, by the first part of the problem. Therefore, the unit sphere contains the countable dense set . By the definition of the norm, we can find the points of norm such that . We will prove that the set is a dense, countable set in . (where means that we take all linear combinations with rational coefficients. Then is a subspace of .) If not, there exists a functional of norm which vanishes on the closure of , and is not identically zero.

Then we have for every . This contradicts the separability of . Therefore the closure of is and is separable.

This shows us right away that , since the latter is separable, and is not.

For the proof of separability of subset you may use such a property. A normed space is separable iff there is no uncountable subsets with pairwise big distance between its points (i.e. all pairwise distances are greater some fixed positive constant). If you assume such a property then the proof of separability becomes obvious. As for the proof of the first statement, the most hard part is to prove that the absence of such uncountable subsets imply separability. You can prove this using Zorn’s lemma. It is easy to see that proposed approach is an overkill. But I think it is very interesting to know such a characterization of separability.

Thank you for your comment. I’ll try and develop a new solution from your suggestions.

Do you know where to find the proof of this characterization of seperability? Seems interesting but I can’t figure it out.

I’ll think about it and I will post a solution these days. It seems interesting. 🙂

I posted a proof at the following link: https://mathproblems123.wordpress.com/2013/09/15/characterization-of-separability/

Your second solution is wrong.

First) It’s clear that F can’t be dense in X, since ||x||=1 for all x in F.

Second) F isn’t a vectorial subspace of X, and then, you can’t apply the Hahn-Banach consequence that you are using.

Greetings

Thank you for your observation. I corrected the proof.