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Does the weak-star topology on the dual of a separable Banach space make the dual completely regular under weak-star topology?

So I have come to the stage in a proof where if I could show this, then I would be done!

In case you are interested in the original problem. That is to show that a weak-star closed subset of the unit ball $B'$ in the dual space, is a Z-set in $B'$

  • Note, this may be true in a much less general case. So please let me know if so! –  Apr 06 '13 at 01:35

2 Answers2

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This is easy to see, as the weak-star-topology is a product topology (this is usually seen in the proof of the Banach-Alaoglu theorem). In fact, if $X$ is your Banach space over the scalars $\mathbb{K}$, then the map $\Phi:X^\star \longrightarrow \prod_{x\in X} \mathbb{K}$ defined by $$ \Phi(x^\star) = (x^\star(x))_{x \in X} $$

is a continuous and open endomorphism. As the image space is completely regular, $X^\star$ is too.

Vobo
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Yes. Indeed, every Hausdorff topological group is completely regular. (In this case, the group operation is the vector space addition, so we actually have an abelian group.)

I just learned this, but it seems to be a well-known consequence of the Birkhoff-Kakutani theorem. There is a proof at Corollary 3.0.7 of these notes.

Nate Eldredge
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  • Thanks for this comment. I am not that familiar with topological groups, so please excuse me if this is well known too! But how can we make use of this fact for Hausdorff topological groups, when there seems to be no algebraic structure? –  Apr 06 '13 at 10:30
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    In fact, every $T_0$ topological group is completely regular. – tomasz Apr 06 '13 at 11:43
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    @rustyracketman: the dual of a Banach space is a vector space; in particular, it is an abelian group. – tomasz Apr 06 '13 at 11:56