Let $X$ be a real Banach space. Apart from the norm topology, we can consider the following weak topologies on $X$:

  • the weak toplogy, defined as the initial topology with respect to $X^*$. In other words, it is the coarsest topology for which all $f\in X^*$ are continuous.
  • the weak sequential topology, which is essentially the topology induced by weak convergence. More precisely, we call a set closed if it is weakly sequentially closed, and this induces a topology.

It is easy to see that the weak topology is the weaker of the two (since every $f\in X^*$ is weakly sequentially continuous). Moreover, it is well known that a weakly sequentially closed set is not necessarily weakly closed. However, the picture is not so clear when it comes to compactness:

By the Eberlein-Smulian theorem, weak compactness coincides with weak sequential compactness. However, it is important to note that weak sequential compactness means sequential compactness, not compactness in the weak sequential topology (!!). In particular, this raises the following question:

What does ordinary compactness (i.e., covering compactness) look like in the weak sequential topology? Is it equivalent to weak (sequential) compactness?

(Also: if these are not equivalent in general, what about the case of convex sets?)

  • $\begingroup$ A convex set is closed if and only if it is weakly closed and if and only if it is weakly sequentially closed. $\endgroup$
    – Dirk
    Nov 3, 2017 at 12:10
  • $\begingroup$ One does not need idempotence of sequential closure in a topological space $X$ to obtain a sequential topology. One simply takes the complements of sequentially closed sets to be the open sets. This produces a topology with no conditions on $X$'s original topology. $\endgroup$ Nov 3, 2017 at 17:45
  • $\begingroup$ The comment that prompted my earlier comment has been deleted, making it look silly. Someone said that idempotence of sequential closure was necessary. $\endgroup$ Nov 3, 2017 at 21:44

2 Answers 2


It seems that in Banach spaces weakly compact subsets remain compact in the weak sequential topology. This follows from three facts:

  1. By a well-known Eberlein Theorem, weakly compact sets in Banach spaces are Frechet-Urysohn (and hence sequential) in the weak topology.

  2. For a sequential space $X$ the sequetial modification of its topology (i.e., the family of all sequentially open sets) coincides with the original topology;

  3. For a sequentially closed subset $K$ of a topological space $X$ the sequential modification of the subspace topology of $K$ coincides with the subspace topology inherited from the sequential modification of the topology of $X$.

  • $\begingroup$ Thanks for your answer. Just a comment: why do so few (if any) textbooks mention the weak sequential topology? It seems to be a (slightly) stronger topology than the weak topology but with the same notion of compactness. $\endgroup$ Nov 5, 2017 at 11:54

Taras' answer seems perfectly fine to me, but I'm adding a second answer which is a little more "elementary" (despite being fundamentally the same).

The key observation is that, for subsets of weakly (sequentially) compact sets, the notions of weak closedness and weak sequential closedness coincide. To see this, let $C$ be weakly compact and $A\subseteq C$ weakly sequentially closed. Then $A$ is weakly sequentially compact (since $C$ is), hence weakly compact (by the Eberlein-Smulian theorem), hence weakly closed. The converse implication (weak closedness $\Rightarrow$ weak sequential closedness) holds trivially.

It follows that, if $C$ is weakly compact, then the weak and weak sequential topologies (or, more precisely, their induced subspace topologies) coincide on $C$. This yields the result.


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