Weakly compact set

A subset of a Banach space is weakly compact when it is compact in the weak topology; it is relatively weakly compact when its weak closure is weakly compact.

A bounded sequence in the space of continuous functions on a compact space that has a discontinuous pointwise limit cannot lie in a weakly compact set. Compactness would give a weakly convergent subnet, and continuous point-evaluation functionals would identify its continuous limit with the prescribed pointwise limit. No assumption of sequential compactness is needed.

Coordinate evaluations on the l-infinity sequence space form a countable separating family, so a weakly compact set $K$ is metrizable by countable separating family metrizes a weakly compact set. Choose a countable weakly dense subset $D$. Then $K$ lies in the weak closure of $\mathrm{conv}D$, which equals its norm closure by Mazur theorem. Rational convex combinations make that norm closure separable, and every subset of a separable metric space is separable. The weakly dense subset itself need not be norm dense in a nonconvex $K$.

Every weakly compact set in a normed vector space is norm bounded, even if the space is incomplete. Its canonical embedding into the bidual is pointwise bounded on the continuous dual space; the Uniform boundedness principle on that Banach dual gives the norm bound.

A set is weakly sequentially compact when every sequence in it has a subsequence that converges weakly to a point of the set.

For a subset of a Banach space, relative weak compactness is equivalent to every sequence having a weakly convergent subsequence. In particular, weak compactness and weak sequential compactness coincide.

The closed convex hull of a weakly compact subset of a Banach space is weakly compact.

The other standard formulation says that a convex subset $C$ of a Banach-space dual is weak-star closed if its intersection with every norm-closed ball is weak-star closed.