I am looking at Murphy's book "$C^*$algebras and operator theory", in the section on abelian Von Neumann algebras (end of chapter 4). There, it is explained that any (unit containing) abelian Von Neumann algebra (say $A$) on a separable Hilbert space (say $H$) with a cyclic vector can be identified via a unitary to an $L^\infty(X,\mu)$ for a second countable compact space $X$.

What I don't understand is the explanation for second countability; namely, using the notations there, why there is a separable $C^*$algebra $B$ contained in $A$ which is strongly dense in $A$? I have looked at the remarks at the beginning of the section, and looked in the book, but I don't see the reason here.

Thanks for any help.


As Murphy says, the unit ball in $B(H)$ is separable and metrizable in the strong topology (see Remark 4.4.2 on pages 133-4 for details). Since a subspace of a separable metric space is separable, the unit ball in $A$ is separable in the strong topology. Now just take a countable strongly dense subset of the unit ball of $A$ and let $B$ be the C*-algebra it generates. Since the subset was countable, $B$ will be separable, and $B$ is strongly dense in $A$ since its strong closure contains the entire unit ball of $A$.

  • $\begingroup$ Thank you for your answer, I see it clearly now ! $\endgroup$ – John Jan 27 '16 at 12:43

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