Proof. (1): Let $\phi \in \Omega(A)$. By Proposition 33.7.2, $\norm{\phi}_{A^*}= \dpn{1, \phi}{A}= 1$. Thus $\phi$ is a state by Theorem 34.9.2, and $\Omega(A) \subset S(A)$.

Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\phi = (1 - t)\psi + t\rho$, then for each $x \in \ker(\phi)$, $x^{*}x \in \ker(\phi)$ as well. As $t \ne 0$, $x^{*}x \in \ker(\psi)$ and $x^{*}x \in \ker(\rho)$. By the Cauchy-Schwarz inequality,

Likewise, $\dpn{x, \rho}{A}= 0$ as well. Hence $\ker(\psi), \ker(\rho) \supset \ker(\phi)$. Thus there exist scalars $\alpha, \beta \in \complex$ such that $\phi = \alpha \psi = \beta \rho$. However, since $\phi, \psi, \rho \in S(A)$, $\alpha = \beta = 1$, and $\phi = \psi = \rho$. Therefore $\phi$ is a pure state.

(2): Using the Gelfand-Naimark Theorem, identify $A$ with $C(\Omega(A); \complex)$ and $S(A)$ as Radon probability measures on $\Omega(A)$.

Let $\cm = \bracs{t\mu|\mu \in S(A), t \in [0, 1]}$. By Proposition 23.7.4, the extreme points of $\cm$ are the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, and possibly $0$. For any $\mu \in S(A)$, $\nu, \rho \in \cm$, and $t \in (0, 1)$, $\mu = (1 - t)\nu + t\rho$ implies that $\nu(\Omega(A)) = \rho(\Omega(A)) = 1$, and $\nu, \rho \in S(A)$ as well. Thus the extreme points of $S(A)$ are exactly the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, which correspond to $\Omega(A)$ itself.$\square$