I know that locally compact Hausdorff topological vector space must be isomorphic to $\mathbb{C}^d$ or $\mathbb{R}^d$ for some $d\in \mathbb{N}$.
However the Banach Alaouglu theorem says that the closed unit ball of $X^*$ is compact with respect to the weak-* topology, which seemed to me at first sight to mean that it is locally compact. But I know that this space with the weak-* topology is also Hausdorff, and allegedly should mean that $X^*$ with the weak-* topology is finite dimensional.
My question is then why is the closed unit ball in $X^*$ not a compact neighbourhood of $0$?
I found this thread saying that exactly, but I do not understand why. Is the unit ball of $X^*$ not a neighbourhood of $0\in X^*$ with the weak-* topology, which is contained in the aforementioned set?
Best Answer
A basic neighbourhood of $0$ (functional) in the weak-star topology is a finite intersection of sets of the form $O(x,r):=\{f \in X^\ast: |f(x)| < r\}$ with $r>0$ and $x \in X$. This only restricts a functional in such a neighbourhood on finitely many $x \in X$, so if we'd have $\bigcap_{i=1}^n O(x_i,r_i) \subseteq B$ where $B$ is the closed unit ball (in norm!!) in $X^\ast$, it's easy to find some functional $f$, by Hahn-Banach, that is $0$ on all finitely many $x_i$ and (say) $100$ on some $x \notin \{x_1,\ldots,x_n\}$ with $\|x\| = 1$ (we need that $X$ is infinite-dimensional here). Then $\|f\|\ge 100$, so $f \notin B$ while $f \in \bigcap_{i=1}^n O(x_i,r_i)$, contradiction.
So no basic neighbourhood of $0$ (in the weak-star topology) lies inside $B$ and so $0 \notin \operatorname{int}(B)$ as you already suspected to be the case.