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 $s_i \to^w x.$

  • $\begingroup$ @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. $\endgroup$ Jan 10 at 16:06
  • $\begingroup$ @SeverinSchraven I erroneously saw an "only if" in the OP... $\endgroup$ Jan 10 at 16:08
  • $\begingroup$ @DavidMitra Makes perfect sense. Also I think it is a good thing to stress that sequences are in general enough. $\endgroup$ Jan 10 at 16:13
  • $\begingroup$ There are weakly dense sets that are not weakly sequentially dense, though. As contained here, I believe. $\endgroup$ Jan 10 at 16:16
  • $\begingroup$ @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. $\endgroup$ Jan 10 at 18:58

1 Answer 1


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.


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