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I know that weak* topology is the weakest topology so that $Jx$ is continuous for $\forall x\in X$, where $J$ is the isometry from $X$ to $X''$. But what exactly is this topology? What is the open set in general look like?

And moreover, I want to prove another topology induced by metric is exactly the weak* topology. How can I prove this topology is weaker than weak* topology, so that it is exactly the weak* topology?

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    @Chris Thank you so much2012-05-02

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Weak*-open (non-empty) sets contain infinite-dimensional subspaces, so in this sense they are huge.

Roughly, we have less open sets, hence more compact setes. This is reflected in the Banach-Alaoglu theorem which says that a polar set of any nbhd of the origin in $X$ is weak* compact in $X^*$.

You work with Banach spaces with separable duals. In this case, the dual ball (which is weak*-compact) is metrisable. Thus, you cannot find any other compact Hausdorff topology (in particular, any other metric giving topology which is compact) which is weaker that the (relative) weak* topology since compact Hausdorff topologies are minimal.

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    You don't need the dual to be separable, only the space itself. This guarantees the metrizability of the unit ball of the dual space.2012-05-05