# State space is weak* compact

I'm trying to convince myself that the state space $S(A)$ of a unital $C^*$-algebra is weak* compact. I've proven that $S(A)$ is convex, and I feel that this should allow me to conclude weak* compactness. However, after awhile, I don't see how this could be the case. Can someone point me in the right direction?

-
Use the Banach–Alaoglu theorem. Now you only need to prove that $S(A)$ is weakly* closed.
Elaboration: The weak* topology is, by definition, the weakest topology on $A^*$ for which every bounded linear functional of the form $\psi\mapsto\psi(a)$, with $a\in A$, is continuous. In particular, applying this to the unit element $e$, we conclude that $\{\,\psi\in A^*\colon \psi(e)=1\,\}$ is weakly* closed. The state space is just the intersection of this hyperplane with the unit ball of $A^*$, which is also weakly* closed (which follows from compactness, but more easily from Hahn–Banach).
Thanks, Harald. I thought about that, but I do not easily see why $S(A)$ is weak* closed. It seems obvious to everybody but me. Could you elaborate a bit on it? I'm having a hard time understanding what is a weak* closed set. –  ragrigg Feb 22 '13 at 8:18
Thanks a lot. I was missing the obvious: the first sentence in your elaboration. By the way, what goes wrong when you don't have a unit? I'm thinking in the non-unital $C^*$-algebra of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ that vanish at infinity. I clearly see that your argument would not work anymore, but I don't see why we will have that $S(A)$ is not compact. Any insight on that one? –  ragrigg Feb 22 '13 at 9:53
@ragrigg In the example you mention, just take point evaluation at $n$ and let $n\to\infty$. You get a sequence of states converging weakly* to the zero functional. What happens in a general non-unital C*-algebra, I am not so sure. –  Harald Hanche-Olsen Feb 22 '13 at 15:47