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A von Neumann algebra is countably decomposable if every family of mutually orthogonal nonzero projections is countable. Even a singly-generated von Neumann algebra need not be countably decomposable; for instance, consider multiplication by $x$ on the nonseparable Hilbert space $l^2[0,1]$. However, every self-adjoint element is approximated in norm by elements of finite-dimensional (and hence countably decomposable) subalgebras.

Let $(x_n)$ be a sequence of positive elements of an arbitrary von Neumann algebra $\mathcal{M}$ which increase weak* to $1$ and let $\epsilon > 0$. Can we find a countably decomposable subalgebra $\mathcal{M}_0$ and a sequence $(y_n)$ in $\mathcal{M}_0$ such that $\|x_n - y_n\| \leq \epsilon$ for all $n$?

I will accept an answer in the case where $\mathcal{M} = B(H)$ (for nonseparable $H$, of course).

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  • $\begingroup$ Can you treat the case when $H= \ell_2(\{0,1\}^\mathbf N)$ and $x_n$ is the indicator function of $\{ \omega \in \{0,1\}^\mathbf N, \omega_n=0\}$? $\endgroup$ – Mikael de la Salle Jul 7 '16 at 23:40
  • $\begingroup$ Yeah, good question. It doesn't seem likely, does it? So, I am going to respond by insisting on my extra condition that the $x_n$ increase to $1$. $\endgroup$ – Nik Weaver Jul 8 '16 at 0:06

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