Proof. By Theorem 34.6.1, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^{*}$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the Stone-Nakano Theorem implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.$\square$