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Recall:

Definition 2.1.2 If $A$ is a C*-algebra and $N$ is a von Neumann algebra, a map $\theta:A \rightarrow N$ is called weakly nuclear if there exist c.c.p. maps $\phi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ and $\psi:M_{k(n)}(\mathbb{C}) \rightarrow N$ such that $\psi_{n} \circ \phi_{n} \rightarrow \theta$ in the point-ultraweak topology.

Definition 2.3.3 A con Neumann algebra $M$ is called semidiscrete if the identity map $id_{M}:M \rightarrow M$ is weakly nuclear.

If $A$ and $B$ are both semidiscrete, then $A\oplus B$ is also semidiscrete?

Addition:

When I prove this question above, I met with a problem. That is,

If $T_{i}$ and $S_{i}$ are point-weak convergent to $T$ and $S$ respectively, then how to verify $T_{1}\oplus T_{2}$ is point-weak convergent to $T\oplus S$?

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Yes. You can factorise the identity map on $A\oplus B$ in the point-ultraweak topology taking direct sums of the respective approximations for $A$ and $B$. Then you pretend that $M_{k_A(n)}$ and $M_{k_B(n)}$ live in some bigger full matrix algebra and you are done. In other words, you can replace the matrix algebras $M_{k(n)}$ in the definition of semidiscreteness by arbitrary finite-dimensional C*-algebras.

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  • $\begingroup$ Yeah, I did like this and I met with a problem in the proof. Please see the addition of question above (I have edited my question). $\endgroup$ – Yan kai Apr 2 '14 at 1:42

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