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In the late 70s, Cuntz and Behncke had a paper

H. Behncke and J. Cuntz, Local Completeness of Operator Algebras, Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100

the about the following question. Let $A$ be a $C^\star$-algebra and let $B \subset A$ be some $*$-subalgebra. Let $B$ be dense and such that every maximal abelian subalgebra of $B$ is norm-closed. Is it true that $B=A$? They proved this in various cases.

I want to ask a related question in the von-Neumann-setting. More precisely: Let $A$ be a von Neumann algebra and $B \subset A$ be some ultra-weakly dense $\star$-subalgebra such every MASA of $B$ is ultra-weakly closed in $A$. Is it true that $B=A$? A result of Gert Pedersen implies that once $B$ is a $C^\star$-algebra, then $B=A$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that $B=A$ if $A$ has no $II_1$-part.

Question: Is it true for every von Neumann algebra?

or a little more modest

Question: Is it true for the hyperfinite $II_1$-factor?

Another strenghtening of the the assumption (which could help) would be the following:

Question: Let $A$ be a $II_1$-factor and $B \subset A$ be a ultra-weakly dense $\star$-subalgebra such that for every hyperfinite subalgebra $R \subset A$ one has that $R \cap B$ is ultra-weakly closed in $A$. Is it true that $B=A$?

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If I understand correctly, you need to add a density hypothesis in your question? –  Martin Argerami Aug 9 '10 at 0:42
    
Corrected, thanks. –  Andreas Thom Aug 9 '10 at 5:04
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