# Maximal abelian von Neumann algebra and cyclic vectors

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!

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.