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Given a Banach space the only way i've seen to show that it is not separable is to show that there is a more than countable set $A$ and a costant $c>0$ such that $|a_1-a_2|>c, \forall a_1 \neq a_2 \in A$(in this way you show that $l^{\infty}$ is not separable). My question is: is it true the opposite implication? That is

Question: Given X a non separable Banach space, is it true that there is $A$ more than countable such that there's $c>0$ so that $|a_1-a_2|>c, \forall a_1 \neq a_2 \in A$?

Thanks! Bye

Jimmy1790
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2 Answers2

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Let $X$ be a non-seperable Banach space, and $\epsilon \in (0,1)$. Construct by induction a sequence $(x_\alpha)_{\alpha < \omega_1}$ as follows:

  • Given $\alpha< \omega_1$, the Banach space $U_\alpha := \overline{\def\span{\mathop{\rm span}}\span \{x_\beta\mid \beta < \alpha\}}$ has a countable total set as is hence separable. So $U_\alpha \ne X$. By the lemma of Riesz, there is an $x_\alpha \in S_X$ (the unit sphere) with ${\rm dist}(x_\alpha, U_\alpha) \ge 1-\epsilon$.

Then $(x_\alpha)_{\alpha < \omega_1}$ has $\|x_\alpha - x_\beta\|\ge 1-\epsilon$ for $\alpha \ne \beta$ and is uncountable.

martini
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Your condition gives you an ability to find an uncountable collection of pairwise disjoint open sets. You can get this collection uniformly by taking balls of diameter of $c / 2$. The existence of such collection is equivalent to non-separability in metric spaces. So your question reduces to wheather one can find (in Banach space) an uncountable collection of paiwise disjoint open sets which is not uniform as in your condition.

user87690
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