# Dense and weakly dense, sequentially speaking.

Let $$X$$ be a normed space and $$X'$$ its dual. The weak topology $$\sigma_w$$ is defined to be the smallest topology such that all functionals in $$X'$$ are continuous. Then we may speak of weak convergence: $$x_i \to^w x$$ if $$\forall l \in X': l(x_i) \to l(x)$$.

Is the following correct?

We say that a set $$S \subset X$$ is dense if $$\overline{S} = X$$. In other words, given any $$x \in X$$, we can find a sequence $$s_1, s_2, ...$$ such that $$s_i \to x$$. Then we can say the same thing about weakly dense? $$S$$ is weakly dense in $$X$$ if for any $$x \in X$$, there is a sequence $$s_1, s_2, ..$$ such that $$s_i \to^w x.$$

• @DavidMitra I agree with your comment, except for the first word. Don't we have that the sequential closure in a Hausdorff space is contained in the closure? Thus, if the sequential closure is already everything, so would the actual closure. Maybe I am missing something. Jan 10 at 16:06
• @SeverinSchraven I erroneously saw an "only if" in the OP... Jan 10 at 16:08
• @DavidMitra Makes perfect sense. Also I think it is a good thing to stress that sequences are in general enough. Jan 10 at 16:13
• There are weakly dense sets that are not weakly sequentially dense, though. As contained here, I believe. Jan 10 at 16:16
• @DavidMitra Of course I meant not enough in my previous comment. Indeed I don't think the reverse direction of what the OP is asking holds true. Jan 10 at 18:58

In a normed space, which is metric hence a sequential space, we indeed have that $$S$$ is dense iff for all $$x \in X$$ some sequence from $$S$$ converges to $$x$$. The set of convergent sequences from $$S$$ is called the sequential closure of $$S$$, $$[S]_{\text{seq}}$$. And for the strong (norm) topology we have $$\overline{S}=[S]_{\text{seq}}$$, while this need not hold for the weak topology or the weak-* topology. See this paper for an historical overview.