If that can't be achieved, what if the Banach space is reflexive?
Asked
Active
Viewed 163 times
1
-
2Take a sequence $x_n$ of linearly independent vectors; consider the closure of their linear span. – Dec 30 '15 at 04:46
-
How could you prove the closure is seperable? – user219967 Dec 30 '15 at 05:02
-
1As a separable dense subset, take linear combinations with rational coefficients. – Dec 30 '15 at 05:05
-
I don't think the subset of linear combinations with rational coefficients is countable… – user219967 Dec 30 '15 at 05:19
-
2It is. I emphasize: finite linear combinations. – Dec 30 '15 at 05:20