# If $X$ is separable space then $X^∗$ is separable in all topologies $\tau$ such that $(X^∗,\tau)^∗ =X$?

Let $$(X,\|.\|_{X})$$ be a separable Banach space and the associated dual space is denoted by $$X^*$$. By $$w^*$$ we shall indicate the weak$$-*$$ topology on $$X^*$$.

Let $$B_{X^∗}= \{x^∗ \in X^∗ : \|x^∗\|_{X^∗}\leq 1\}$$. Since $$X$$ is separable, the set $$B_{X^∗}$$ furnished with the relative $$w^∗-$$topology is compact (by the Alaoglu theorem) and metrizable (see Theorem I.5.85). Note that $$X^*=\bigcup_{n}{nB_{X^*}}$$ hence $$X^∗_{w^∗}$$ (the space $$X^∗$$ furnished with the $$w^∗-$$topology) is separable.

Can we say that : $$X^∗$$ is separable in all topologies $$\tau$$ such that $$(X^∗,\tau)^∗ =X$$?