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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 $vN(A)_f$ is dense in $vN(A)$ for 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 $A$.

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 $A$.

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 $bN(A)$ 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 $vN(A)_f$ is dense in $vN(A)$ for 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 $A$.

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 $vN(A)_f$ is dense in $vN(A)$ for 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 $A$.

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 $bN(A)$ 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 $vN(A)_f$ is dense in $vN(A)$ for 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 $i$, $d_2(x_i,K) < \varepsilon$.

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 $bN(A)$ 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 $vN(A)_f$ is dense in $vN(A)$ for 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 $A$.