TITLE: Is weak-* the strongest topology s.t. the unit ball is compact?
QUESTION [5 upvotes]: Let $B$ be a Banach space, $B^*$ be the dual of $B$ and $\mathcal B$ be the closed unit ball in $B^*$.
By the Banach-Alaoglu theorem, $\mathcal B$ is closed in the weak-* topology.
My question is whether in general the weak-* topology on $B^*$ is the strongest topology such that this assertion holds?
Thank you

REPLY [4 votes]: No. The compactness of unit ball depends only on the open sets intersecting the unit ball. You can add lots and lots of new open sets far from the unit ball without making the unit ball noncompact.
For instance, introduce a new topology on $B^*$: a subset $U \subset B^*$ is open if and only if there's some set $V$ open in the weak-* topology such that $U \cap \mathcal{B} = V \cap \mathcal{B}$. This condition ensures that the induced topology on $\mathcal{B}$ is the same as topology induced from weak-* topology. Now, since these topologies are the same, $\mathcal{B}$ is compact in this new topology. The complement of the $\mathcal{B}$ in $B^*$ has a discrete topology, though. 
Probably you were thinking of some topology that's related to the Banach space structure?