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A Banach space is a topological group under addition. The dual is a topological group under the weak$^*$ topology. The weak$^*$ topology is weaker than the operator norm topology, so is it first-countable? It is also Hausdorff. (Seems intuitively obvious, might require the Hahn-Banach Theorem to make rigorous). So by the Birkhoff-Kakutani Theorem it should be metrizable. But I believe this is only the case for the finite-dimensional case. Where is my reasoning wrong?

Yiorgos S. Smyrlis
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Daron
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    "The weak* topology is weaker than the operator norm topology, so is first-countable" is, except in the finite-dimensional case, wrong. – Daniel Fischer Jan 11 '14 at 16:44
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    First countability doesn't transfer to stronger or weaker topologies. Discrete topology and trivial topology are both always first countable and any topology is between the two. – tomasz Jan 11 '14 at 16:52

1 Answers1

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If a topological space is metrizable, then it has to be first countable.

It turns out that, if $X$ has uncountable base (as a linear space), then the weak$^*$ topology of $X^*$ is not first countable. See

The weak$^*$ topology on $X^*$ is not first countable if $X$ has uncountable dimension.

But, even if the dimension of a normed space is $\aleph_0$, then its dual shall coincide with the dual of the completion of $X$, which will have dimension $2^{\aleph_0}$, and hence the weak$^*$ topology of $X^*$ will not be first countable, as well.

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Yiorgos S. Smyrlis
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    Is the weak* topology on $X^$ induced by the completion of $X$ the same as the weak topology on $X^*$ induced by $X$? – David Mitra May 29 '14 at 22:20
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    If and only if $X$ is complete. I'm sure that was a rhetorical question by @DavidMitra, but just in case somebody doesn't know: if $Y$ is a (linear) subspace of the completion of $X$ such that $\lambda\lvert_Y = \mu\lvert_Y \implies \lambda = \mu$ for $\lambda,\mu\in X^\ast$, then the dual space of $X^\ast$ with the weak* topology induced by $Y$ is $Y$, so no two such subspaces induce the same weak* topology on $X^\ast$. – Daniel Fischer May 30 '14 at 23:44
  • Please explain more about when "the dimension of a normed space is ℵ 0 " part. Reading above result, I can not understand it or give me a reference about it. – niki Jan 08 '15 at 08:13
  • Dear @DanielFischer or others also who could give me a hand : what happens if $X$ is not Banach? I am not sure why this is neccessary (or convenient) in the above argument. Thank you. –  Apr 09 '22 at 19:21
  • What happens if $X$ is not Banach? Can't we say also that then its dual has an uncountble dimension? @DavidMitra Thank you. –  Apr 10 '22 at 13:21
  • Dear @Yiorgos, could you please give a reference of where can I read that the dimension of continuous dual is also always strictly bigger than that of the original space? And what happens if $X$ is not Banach? – user1110 Apr 15 '22 at 23:29
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    For the sake of correcting records, the claim in the final paragraph is false. The argument (here and in the linked answer) does not work as the weak$^\ast$ topologies on $X^\ast$ and $\hat{X}^\ast$ (where $\hat{X}$ is the completion of $X$) do *not* coincide, since the former has dual space $X$ while the latter has dual space $\hat{X}$. Indeed, the claim is simply wrong, because the weak$^\ast$ topology on $X^\ast$ for a countably infinite-dimensional $X$ *is* metrizable - its topology is generated by countably many seminorms, corresponding to a basis of $X$. – David Gao Aug 14 '24 at 15:22