Why does anyone care about these topologies? The short answer is that in the weak ∗ topology, closed unit ball in X^{′} is compact. This is not true in the normal topology. This wonderful result is the Banach Alaoglu theorem. First recall the notion of the product topology, and the Tychonoff theorem, Theorem 22.3.3 on Page 1546 which are stated here for convenience.
Definition 23.5.2 Let I be a set and suppose for each i ∈ I,

where B_{i} = X_{i} if i≠j and B_{j} = A. A subbasis for a topology on the product space consists of all sets P_{j}
The Banach Alaoglu theorem is as follows.
Theorem 23.5.4 Let B^{′} be the closed unit ball in X^{′}. Then B^{′} is compact in the weak ∗ topology.
Proof: By the Tychonoff theorem, Theorem 23.5.3

is compact in the product topology where the topology on B

for some x,y ∈ X. However, if g is close enough to f at the three points, x + y,x, and y, the above inequality will hold for g in place of f. In other words there is a basic open set containing f, such that for all g in this basic open set, g
Sometimes one can consider the weak ∗ topology in terms of a metric space.
Theorem 23.5.5 If K ⊆ X^{′} is compact in the weak ∗ topology and X is separable in the weak topology then there exists a metric, d, on K such that if τ_{d }is the topology on K induced by d and if τ is the topology on K induced by the weak ∗ topology of X^{′}, then τ = τ_{d}. Thus one can consider K with the weak ∗ topology as a metric space.
Proof: Let D = {x_{n}} be the dense countable subset in X. The metric is

where ρ_{xn}

and f

valid whenever x,y ≥ 0. Therefore this is a metric.
Thus there are two topological spaces,
Now suppose U ∈ τ. Is U in τ_{d}? Since K is compact with respect to τ, it follows from the above that K is compact with respect to τ_{d} ⊆ τ. Hence K ∖ U is compact with respect to τ_{d} and so it is closed with respect to τ_{d}. Thus U is open with respect to τ_{d}. ■
The fact that this set with the weak ∗ topology can be considered a metric space is very significant because if a point is a limit point in a metric space, one can extract a convergent sequence.
Note that if a Banach space is separable, then it is weakly separable.
Corollary 23.5.6 If X is weakly separable and K ⊆ X^{′} is compact in the weak ∗ topology, then K is sequentially compact. That is, if {f_{n}}_{n=1}^{∞}⊆ K, then there exists a subsequence f_{nk} and f ∈ K such that for all x ∈ X,

Proof: By Theorem 23.5.5, K is a metric space for the metric described there and it is compact. Therefore by the characterization of compact metric spaces, Proposition 2.5.5 on Page 65, K is sequentially compact. This proves the corollary. ■