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Aug 23, 2019 at 22:25 comment added Robert Furber The weak topology on $E$ defined by a separating vector space of linear functionals $F \subseteq E^*$ is first-countable iff it is metrizable iff $F$ is of countable (Hamel) dimension. So we can get metrizability if "Banach" is relaxed to "normed".
Mar 27, 2014 at 1:32 answer added Samuel G. Silva timeline score: 6
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Jan 14, 2011 at 22:10 vote accept Sudip Paul
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Jan 14, 2011 at 22:10
Jan 14, 2011 at 22:09 vote accept Sudip Paul
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Jan 14, 2011 at 7:05 comment added Theo Buehler @Anthony: No, that's not true. It is true that the unit ball in $X^{\ast}$ is weak$^{\ast}$-metrizable if $X$ is separable, but this does by no means imply that the weak$^{\ast}$-topology on all of $X^{\ast}$ is metrizable. In fact, the weak$^{\ast}$-topology is not even first countable if $X$ is infinite-dimensional.
Jan 14, 2011 at 7:02 answer added Zen Harper timeline score: 14
Jan 14, 2011 at 6:43 history made wiki Post Made Community Wiki by Sudip Paul
Jan 14, 2011 at 6:16 comment added Anthony Quas As far as I know the w* topology on the dual is metrizable as long as $X$ is separable
Jan 14, 2011 at 6:07 answer added Nate Eldredge timeline score: 30
Jan 14, 2011 at 5:31 comment added Nate Eldredge Have you had a look at Counterexamples in Topology?
Jan 14, 2011 at 4:35 history asked Sudip Paul CC BY-SA 2.5