In this note, page 25, the 4-th item of the section "Properties of Weak Topologies" states that

For every net $\{x_\alpha\}_{\alpha\in I}\subset X$, by a theorem in the section on locally convex subsets, we have $\{x_\alpha\}_{\alpha\in I}$ weakly converges to $x$ iff for every $f\in X^*$, $f(x_\alpha)\to f(x)$.

I think I can prove it readily: by the very definition of the weak topology on $X$, it makes each $f\in X^*$ continuous; then just note the convergence of the corresponding nets is just an equivalent characterisation of a continuous map between any topological spaces.

One thing I do not understand very well, however, is why one would have to consider nets instead of just sequences. For what I know, the weak topology is still a vector topology on $X$ (see for example Theorem 3.10, Functional Analysis, W.Rudin), and any vector topology is Hausdorff and the toplogy on the scalar field is, of course, too good to allow any pathological properties, therefore the continuity of a map from $X_w$ (denoting $X$ topologised by the weak topology) to the scalar field should be characterised by the Heine property: namely, if $x_n\to x$ weakly then $f(x_n)\to f(x)$. So I do not see a reason to introduce nets here. Am I missing anything crucial?

Thanks to Economist's answer which spotted that the problem is even in Hausdorff spaces sequential continuity need not imply canonical continuity, and the key condition to make the implication true is the first countability instead of Hausdorff-ness.

  • 3
    $\begingroup$ I would also like to point to Tomek Kania's answer math.stackexchange.com/a/2030014/58577. It shows that in each infinite-dimensional Banach space, there is a sequentially weakly closed set which is not weakly closed. $\endgroup$
    – gerw
    Commented Nov 28, 2016 at 7:48
  • $\begingroup$ @gerw thanks for the useful material. $\endgroup$
    – Vim
    Commented Nov 28, 2016 at 17:40

1 Answer 1


The short answer is that locally convex topologies are not necessarily (and in fact, I believe, typically) first countable. This means that sequences are not necessarily enough to characterise the structure of the space.

For example, in an infinite dimensional topological vector space, there exist nets that converge weakly to zero but are unbounded in norm. Hence, the weak topology on an infinite dimensional topological vector space is strictly weaker than the norm topology. The construction of such a net that I'm familiar with cannot be reduced to a sequence.

Edit: To make previous claims more precise: for any normed space $X$, the weak topology on $X$ is first countable if and only if $X$ is finite dimensional, in which case the norm topology and the weak topology are equivalent. You can prove this using the aforementioned construction of a net converging weakly to zero but is unbounded in norm.

Moreover, the Schur property of $\ell_1$ guarantees that that sequences in $\ell_1$ converging weakly to zero also converge to zero in norm, so there is loss of generality when considering only sequences.

  • $\begingroup$ Would you care to elaborate on your example? $\endgroup$
    – Vim
    Commented Nov 27, 2016 at 3:34
  • $\begingroup$ It's essentially the same as the one you can find here. Of course, it doesn't quite prove that you can't do the same with a sequence, just that this specific construction necessitates the use of nets. It seems to be the standard example, however, so I'd be surprised if there actually is one using only sequences. $\endgroup$ Commented Nov 27, 2016 at 3:37
  • $\begingroup$ thanks for your example. Now I guess perhaps the problem with trying to characterise continuity using only sequences is, as you said, the weak vector topology need not be $C_1$, though it is $T_2$. $\endgroup$
    – Vim
    Commented Nov 27, 2016 at 3:51

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