# Unit ball separable $\Longrightarrow$ Space separable

Given a normed space $X$ and assume it is also a locally convex space in some other topology (e.g. weak or weak* if it's a dual). Assume that the unit ball $B_X$ is separable in this topology. Is it then true that the $X$ is separable in this topology?

I think that this is true. Let $D\subset B_X$ be dense. Then $\bigcup_{n\in\mathbb{N}} n D$ is dense in $X$, right?

My attempt for the proof: Given $x\in X$, then $\frac{1}{N} x \in B_X$ for some $N$ large enough. Now there exists a sequence $(y_n)_n \subset D$ such that $y_n \to \frac{1}{N}x$ as $n\to\infty$. Hence $N y_n \to x$ and clearly $Ny_n \in N D$.

Remark: In my definition of a LCS, the topology is also Hausdorff.

• Your proof looks good to me. – icurays1 May 23 '15 at 20:04
• Does this also work if the topology is not first countable? E.g. weak/weak* topology. – Peter May 23 '15 at 20:05
• Let me get this straight. You said that $(X,\tau)$ (where $\tau$ is not necessairly the topology induced by the norm) is locally convex. Moreover, $(B_X,\tau_{B_X})$ is separable. You want to show that $(X,\tau)$ is separable, am I right? – user228113 May 23 '15 at 20:08
• Yes, exactly. That's what I want to do. – Peter May 23 '15 at 20:09

Your proof is fine. You constructed a countable subset of $E\subset X$, then proved that each $x\in X$ has a sequence $(y_n)\subset E$ such that $y_n\rightarrow x$ in your topology. You use the linearity of $X$ and the separability of $B_X$ to construct $E$ and $y_n$, and the Hausdorff condition (implicitly) to prove that $y_n\rightarrow x$. There's nothing hiding under the covers!