Von Neumann algebras have the following form of interpolation property: let $(x\_n)\_n$ and $(y\_n)$ be increasing and decreasing, respectively, sequences of self-adjoint elements in a von Neumann algebra $M$ such that $x\_n \leqslant y\_n$ for all $n$. Then there is $z$ such that $x\_n \leqslant z \leqslant y\_n$ for all $n$.


Can it be strengthened to the following statement. Let $A$ and $B$ be two countable sets of selfadjoint elements with $A \leqslant B$ (each element of $A$ is dominated by each element of $B$). Is there $z$ such that $A \leqslant \{z\} \leqslant B$?


It is true in the commutative case but I am not sure if it works for matrices though.



Certain interpolation property of von Neumann algebras
It does not hold for matrices. Let $P\_1 := \left[\begin{array}{cc} 1& 0 \\ 0& 0 \end{array}\right]$, $P\_2:= \left[\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right]$, and $X:= \left[\begin{array}{cc} 3 & 2,5 \\ 2,5 & 4\end{array}\right]$, and take the sets $A:=\{P\_1, P\_2\}$ and $B:=\{X, Id\}$. Both $X$ and $Id$ are upper bounds for the set $A$, so the assumptions are satisfied. But any self-adjoint matrix $z$ that is an upper bound for $A$ and a lower bound for $Id$ has to be equal to the identity: indeed, its diagonal entries have to be equal to $1$, and therefore its off-diagonal entries have to be zero. But identity is not a lower bound for $X$, so there is no $z$ separating $A$ and $B$ at all.


