Corollary 37.4.6.label Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a maximal abelian von Neumann algebra, then $A$ admits a cyclic vector.
Proof, [Corollary 22.4, Zhu93]. By Proposition 37.4.2, $A = A'$. By Theorem 37.4.5, $A$ admits a separating vector. By Proposition 37.4.3, this separating vector for $A$ is a cyclic vector for $A' = A$.$\square$
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