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I know the fact that if $A\subset B(H)$ is a maximal abelian von Neumann algebra and $H$ is separable, then $A$ will have a cyclic vector. This result is proved in Conway's book "A course in operator theory", Theorem 14.5.

I wonder if the Hilbert space is not separable, can we get the same result that $A$ still has a cyclic vector? In Conway's proof, the fact that $A$ has a cyclic vector strongly depends on the separability of the Hilbert space. So I guess the answer will be No in non-separable case, but I can't find a counterexample right now.

So my question is, $A\subset B(H)$ is a maximal abelian von Neumann algebra and $H$ is non-separable, do we know that $A$ has a cyclic vector? If so, how to prove it? If not so, do we have any counterexample?

Any help will be truly grateful!

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Indeed separability can't be avoided. Suppose $H=\ell^2(S)$, where $S$ is an uncountable set. If $A$ is the algebra of all diagonal operators on $H$ then $A$ is maximal abelian but doesn't have a cyclic vector. The reason is that any vector $\xi=(\xi_i)_{i\in S}$ can only have countably many nonzero components, that is $$N:=\{i\in S: \xi_i\neq0\}$$ is countable. The cyclic space generated by $\xi$ will be contained in $\ell^2(N)$, so $\xi$ can't be cyclic.

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