If $X$ is a Banach space then it's quite straightforward to show that for $A$ a subspace we have $\bar{A} = {(A^{\circ})}_{\circ}$ and so if $A$ is finite dimensional then $A = {(A^{\circ})}_{\circ}$.
I'm wondering what we can say about $B \subset X'$. Is it true that $\bar{B} = {(B_{\circ})}^{\circ}$? I guess if $X$ is reflexive we can use the Hahn-Banach theorem to get suitable elements in $X'' = X$.
I think if we take $B$ to be the set of eventually terminating sequences in $l^{\infty}$ then ${(B_{\circ})}^{\circ} = l^{\infty}$ and so it's not true in general because the terminating sequences are not dense in $l^{\infty}$.
Does it perhaps hold if $B$ is a finite dimensional subspace of $X'$?
Thanks for any help
If $B\subset X'$ then that double annihilator is the weak-* closure of $B$.