Rudy the Reindeer
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 Nov 16 revised Prove that $X^\ast$ separable implies $X$ separable rolled back to a previous revision Nov 16 comment Prove that $X^\ast$ separable implies $X$ separable And is it okay to say "dense" instead of "norm dense"? Nov 16 comment Prove that $X^\ast$ separable implies $X$ separable Hi Brian. Thanks for your helpful answer! I edited the case where I try to show that $x_n^\ast (x_n)$ does not equal $0$. Is case (i) now correct? Nov 16 revised Prove that $X^\ast$ separable implies $X$ separable Trying to slightly improve the writing in this question. Nov 16 accepted Prove that $X^\ast$ separable implies $X$ separable Nov 15 comment Prove that $X^\ast$ separable implies $X$ separable @t.b.: But in the lecture notes it doesn't say continuous. How do I know you're right? Nov 15 comment Prove that $X^\ast$ separable implies $X$ separable @AsafKaragila: Then I mean the algebraic dual, I think. Nov 15 comment Prove that $X^\ast$ separable implies $X$ separable True. I wasn't actually doing it on purpose to annoy you. I'll try to not use symbols anymore from now on. I do remember that you have told me this before, although I only remember one time, not several. Off to read the 2 links in your answer. Nov 15 revised Prove that $X^\ast$ separable implies $X$ separable Typo correction. Nov 15 comment Prove that $X^\ast$ separable implies $X$ separable @kahen: Do I understand correctly: the algebraic dual includes all linear functionals and the topological dual all continuous functionals? Nov 15 comment Prove that $X^\ast$ separable implies $X$ separable @AsafKaragila: I thought that if I could show that there aren't any at all, then there also aren't any continuous ones. Or what do you mean? Nov 15 comment Prove that $X^\ast$ separable implies $X$ separable @kahen: I used the same notation as on my homework sheet. I will do what you suggest whenever I'll be using my own notation. Thanks! Nov 15 comment Prove that $X^\ast$ separable implies $X$ separable @AsafKaragila: If $X$ is a vector space over some field $K$ then $X^\ast$ is the (vector) space of all linear functionals $X \rightarrow K$. Nov 15 revised Prove that $X^\ast$ separable implies $X$ separable Typo correction. Nov 15 asked Prove that $X^\ast$ separable implies $X$ separable Nov 14 accepted A question about regularity of Borel measures Nov 14 comment A question about regularity of Borel measures @t.b.: thanks for the nice answer! Nov 14 comment A question about regularity of Borel measures @t.b.: To show that it's closed under countable union one can use the equivalent definition on Wikipedia. Then for each $i$ there are $F_i \subset A_i \subset G_i$ with $\mu(G_i - F_i)$ small enough. Set $G := \cup_i G_i$ and $F := \cup F_i$ then $F \subset A \subset G$ and $\mu(G - F) \leq \sum_i \mu((G_i - F_i)) < \delta$. Nov 14 revised A question about regularity of Borel measures added 23 characters in body Nov 14 comment Was it an even deal for me? @joriki: You're OCD about the English language : D (and possibly about any other languages you speak) +1 for pointing out a typo.