Proof. Let $\OM'(A)=\OM(A)\cup\curl{0}$ then $\OM'(A)\suf S^{*}$ by Proposition 7.4.10. For each $x,y\in A$, the evaluation

is $\s(A^{*},A)$-continuous thus

which is closed. Thus $\OM'(A)$ is compact by Alaoglu’s Theorem. Since $\curl{0}$ is closed in $\OM'(A)$, $\OM(A)$ is an open subset of the compact space $\OM'(A)$ so it is locally compact. Suppose $A$ is unital then $0$ is isolated in $\OM'(A)$ because $\om(1)=1$ for every $\om\in \OM(A)$ so $\OM(A)$ is closed in $\OM'(A)$; hence compact. Lastly since $A^{*}$ is Hausdorff with $\s(A^{*},A)$-topology and Hausdorff is inherited by subspace we are done.$\square$