Characterization of hyperfinitness by the Effros-Marechal topology 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 $A$.
 A: I believe that the answer is no; in fact, the issue of whether every factor is in the closure of finite-dimensional factors is open (and is equivalent to the Connes embedding conjecture).  I think this is proved in The Effros–Maréchal Topology in the Space of von Neumann Algebras, II by Uffe Haagerup and Carl Winsløw.
A: 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.
