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Separable spaces

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

Prove that if Y is a subset of a separable metric space (X,d) then Y is also separable.

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

Proof: Let X be a separable space and denote by E=\{x_n : n=0,1,...\} a countable dense set of X. We cannot just say that since Y is a subset of X, then some part of E is dense in Y, because it is possible that most of the elements of E to be in X \setminus Y. Therefore, we need another approach.

Then because E is dense, we have that for any n,k \in \Bbb{N} there exists y_{n,k} \in Y such that d( y_{n,k},x_n) \leq \displaystyle\frac{1}{2^k}+d(x_n,Y). Denote \mathcal{D}=\{y_{n,k} : n,k=0,1,...\}\subset Y, and prove that this  is a countable dense set in Y. It is clearly a countable set, as the countable union of countable sets. Take y \in Y and \varepsilon >0. There exists x_j such that d(x_j,y)< \varepsilon/3. So d(x_j,Y)<\varepsilon/3 and for k such that 2^{-k}<\varepsilon/3 we have that

d(y,y_{j,k})\leq d(y,x_j)+d(x_j,y_{j,k})\leq d(y,x_j)+2^{-k}+d(x_j,Y)< 3\cdot \varepsilon/3=\varepsilon, which finishes the proof.

For the second part, let X be a normed vector space and denote by X^\prime its dual space. By Hahn-Banach’s Theorem, for every proper closed subspace Y there exists a linear functional f which vanishes on Y, has norm 1 and for some x_0 \notin Y we have f(x_0)=d(x_0,Y). Suppose now that X^\prime is separable. Then its unit sphere \{f \in X^\prime : \|f\|=1\} is also separable, by the first part of the problem. Therefore, the unit sphere contains the countable dense set E=\{f_n : \|f_n\|=1,\ n=0,1,...\}. By the definition of the norm, we can find the points x_n \in X of norm 1 such that |f_n(x_n)| \geq 1/2. We will prove that the set F=Span_\Bbb{Q}(\{x_n : n=0,1,...\}) is a dense, countable set in X. (where Span_\Bbb{Q} means that we take all linear combinations with rational coefficients. Then \overline{F} is a subspace of X.) If not, there exists a functional f of norm 1 which vanishes on the closure of F, and is not identically zero.

Then we have 1/2 \leq |f_n(x_n)|=|f_n(x_n)-f(x_n)|\leq \|f_n-f\|\|x_n\|=\|f_n-f\| for every n=0,1,... This contradicts the separability of X^\prime. Therefore the closure of F is X and X is separable.

This shows us right away that (\ell^\infty)^\prime \neq \ell^1, since the latter is separable, and \ell^\infty is not.

  1. Norbert
    January 9, 2012 at 8:50 am

    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.

  2. neozt2k2000
    February 7, 2012 at 9:06 pm

    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.


  3. February 7, 2012 at 10:04 pm

    Thank you for your observation. I corrected the proof.

  1. September 15, 2013 at 6:00 pm

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