# Is the weak-star topology on the dual of a Banach space completely regular?

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! – user58514 Apr 6 '13 at 1:35

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.
• In fact, every $T_0$ topological group is completely regular. – tomasz Apr 6 '13 at 11:43