Recall that a topological space is sequential, iff every sequentially closed set is already closed.

Is there an infinite-dimensional Banach space on which the weak topology is sequential?

I already know that the weak topology is not first countable, but (AFAIK) this does not imply that the weak topology is sequential.

On Hilbert spaces, the answer is negative: if $\{e_n\}_{n \in \mathbb{N}}$ is a ONS, then the set $\{\sqrt{n}\,e_n\mid n\in\mathbb{N}\}$ is sequentially weakly closed, but not weakly closed ($0$ belongs to the weak closure).

Edit: On $\ell^1$, the weak topology is also not sequential. This can be seen by using the Schur property of $\ell^1$.


1 Answer 1


No, it cannot be sequential unless $X$ is finite-dimensional. Otherwise, for each $k$ we may pick a subspace $X_k$ of $X$ with $\dim X_k = k$. Moreover, we choose a finite $\tfrac{1}{k}$-net $x_{k,j}$ of the sphere $\{x\in X_k\colon \|x\|=k\}$ in $X_k$ (possible by compactness). Let $S$ be the union of all the nets picked above. We claim that 0 is in the weak closure of $S$.

Indeed, let $U$ be a weakly open neighbourhood of 0. Let $f_1, \ldots, f_n\in X^*$ be norm-one functionals and let $\varepsilon > 0$ be such that $$\{x\in X\colon \max_i |\langle f_i, x\rangle| < \varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $n<k$, there must be $y_k\in X_k$ such that $\langle f_i, y_k\rangle = 0$ for all $i$. Without loss of generality $\|y_k\|=k$. Pick $j$ so that $\|x_{k,j} - y_k\|\leqslant\tfrac{1}{k}$. Consequently, $$|\langle f_i, x_{k,j}\rangle| = |\langle f_i, x_{k,j} - y_k\rangle| \leqslant \|x_{k,j} - y_k\|\leqslant\tfrac{1}{k}<\varepsilon, $$ that is $x_{k,j}\in U$.

This establishes the claim and thus $S$ is not weakly closed.

On the other hand, every weakly convergent sequence in $S$ is bounded, and thus lives only on finitely many points of $S$. Hence, the weak limit belongs to $S$. This yields that $S$ is weakly sequentially closed.

There is a strengthening of this result by Gabriyelyan, Kąkol and Plebanek (see Theorem 1.5 here):

Theorem. Let $E$ be a Banach space. Then the weak topology of $E$ has the Ascoli property if and only if $E$ is finite-dimensional.

  • $\begingroup$ Thank you. I have taken the liberty to slightly edit your proof. I also changed the last paragraph and emphasized that $S$ is sequentially weakly closed. $\endgroup$
    – gerw
    Nov 25, 2016 at 13:44
  • 1
    $\begingroup$ It seems like this argument works even if X is incomplete, i.e., if X is only a normed space. Also, Theorem 3.2 of the paper cited says, for a normed space E, the weak topology of E has the Ascoli property (which is implied by sequential) iff E is finite dimensional. $\endgroup$ Jun 30, 2018 at 17:28

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