Let $A$ be a von Neumann algebra and $A\_\*$ its predual. One can define a topology on the set $vN(A)$of all von Neumann subalgebras of $A$ called the Effros-Marechal topology. It is characterized as the coarsest topology on $vN(A)$ such that for all $\varphi \in A\_\*$ the map
$ B \mapsto ||\varphi\_{|B}|| $
is continous.


Let $vN(A)\_f \subset vN(A)$ be the set of all finite dimensional subalgebras of $A$.


Is it true that the von Neumann algebra $A$ is hyperfinite if and only if $A$ is in the closure of $vN(A)\_f$ in the Effros-Marechal topology?


I believe that this is true at least when $A$ is a $II\_1$-factor because we know that in this case $A$ is hyperfinite if and only if for every $\varepsilon > 0$ and every finite family $x\_i$ of elements of $A$, there existe a finite dimensional subalgebra $K \subset A$ such that for all $i$, $d\_2(x\_i,K) < \varepsilon$.


$d\_2$ being the distance arising from the $||\cdot||\_2$-norm on $A$.



Characterization of hyperfinitness by the Effros-Marechal topology
After some research, I now think this statement is true for any von Neumann algebra $M$.


Indeed, U.Haagerup and C.Winslow give in their paper "The Effros-Marechal topology in the space of von Neumann algebras" various descriptions of this topology. 
In particular, they define for a net $(N\_i)\_{i \in I}$ of von Neumann subalgebras of $M$ a notion of $ \lim\inf $ and $\lim \sup $ such that $N\_i$ converges to a von Neumann subalgebra $N \subset M$ for the Effros-Marechal topology if and only if we have


$ \lim\inf N\_i = N = \lim \sup N\_i $


We have


$\lim\inf N\_i=\{ x \in M \mid \forall U \in n(x), \; \exists i \in I, \forall j\geq i, \; N\_j \cap U \neq 0$ }


where $n(x)$ is the set of $\*$-ultrastrong neighborhoods of $x$.


Now, by definition, $M \in \overline{vN(M)\_f}$ if and only if there exists a net $N\_i$ of finite dimensional $\*$-subalgebras of $M$ such that $N\_i$ converges to $M$. And for this convergence to hold we only need that


$ \lim\inf N\_i = M$


From this we see that $M \in \overline{vN(M)\_f}$ if and only if for every finite family $x\_k \in M$ and every family of neighborhoods $U\_k \in n(x\_k)$, there exists a finite dimensional $\*$-subalgebra $N \subset M$ such that $N \cap U\_k \neq 0$ for all $k$.


But, it is well-known that is a possible definition of hyperfinite von Neumann algebras (see the book of M.Takesaki for example).


Note that if we have $A \in \overline{vN(M)\_f}$, we can **not** conclude that $A$ is hyperfinite because $\overline{vN(M)\_f}$ is in general strictly bigger than $\overline{vN(A)\_f}$. So there is no contradiction with the answer above.


