# The weak-star topology is Completely Hausdorff (in particular Hausdorff).

Let $X$ be a normed space, $X^*$ its dual space, $(X^*, w^*)$ is completely Hausdorff.

Proof: Let $f, g \in X^*$, $f\neq g$ then $\exists x\in X$ such that $f(x)\neq g(x)$ i.e. $\hat{x}(f)\neq \hat{x}(g)$ (here $\hat{x}\in X^{**}$, $\hat{x}(h)=h(x)$ for $h\in X^*$) so $(X^*, w^*)$ is completely Hausdorff.

Note: A top space is Completely Hausdorff when for any given pair of points in it there exists a real valued continuous function that separates them.

The whole thing seems pretty straight forward but it is always good to check.

EDIT: I'm thinking, if the scalar field is complex, we can take either $\text{Re} \hat{x}$ or $\text{Im} \hat{x}$ as the separating continuous function. Also, this function is usually required to map into $[0,1]$. It is sufficient that it maps into $\mathbb{R}$ if the image is bounded though. The whole thing does not bother me because I'm only interested is proving that the unit closed ball $(B_{X^*}, w^*)$ is Hausdorff, which by Banach-Alaoglu is compact (so its image by a real cont function will be bounded). However I asked the question in a more general sense and I am wondering now.

• Looks okay to me. – Brian M. Scott Dec 16 '14 at 2:41
• I'd remark that "every pair of points can be separated by a continuous function" isn't quite the same as "there is a continuous function that separates every pair of points". – Jonas Dec 16 '14 at 2:54
• In general, the weak-star topology will be Tychonoff by virtue of being a $T_0$ vector space topology. – Henno Brandsma Dec 16 '14 at 20:23

The proof is mostly correct, except the step where you note that $\hat x\in X^{**}$: this makes it seem like you are trying to show that the weak topology is completely Hausdorff on a dual space.
That $\hat x\in X^{**}$ is immaterial, what matters here is that $x\in X$.
A different way to see that weak$^{*}$ topology is completely Hausdorff (and actually Tychonoff) - and as a consequence, so is the weak topology on any dual space, because it is finer) is to note that the weak$^{*}$ topology is just the topology of pointwise convergence on $X$.
Since complete regularity is preserved by arbitrary products and by taking a subspace, it follows that any subspace of ${\bf R}^X$ (or ${\bf C}^X$), including $X^*$, is completely regular (Hausdorff).