TITLE: non separable implies an uncountable set with lower bounded distances?
QUESTION [5 upvotes]: 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

REPLY [4 votes]: 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.