direct sum of a collection of Von-Neuman algebras is still Von-Neuman

Question

If {Aα} be a collection of some Von Neuman algebras then their direct sums is still Von Neuman ?

I can prove that if Aα are unital then (⊕ Aα)= (⊕ Aα)" that is because of the fact that (⊕ Aα)'= ⊕(Aα') which implies ⊕ Aα is Von Neuman

but if Aa where not unital then how can i show this statement . In fact if A= (⊕ Aα) I need to show that A is strongly closed

Answer 1

Yes. Let $(A_i)$ ($i\in I$) be a family of von Neumann algebras with preduals $((A_i)_*)$, repsectively. Then the direct sum

$$\Big(\bigoplus_i A_i\Big)_{\ell_\infty(I)} = \{f\colon I\to \bigcup_i A_i\colon f(i)\in A_i\,(i\in I)\text{ and }\sup_{j}\|f(j)\|<\infty\}$$

has a natural predual, that is,

$$\Big(\bigoplus_i (A_i)_*\Big)_{\ell_1(I)} = \{f\colon I\to \bigcup_i (A_i)_*\colon f(i)\in (A_i)_*\,(i\in I)\text{ and }\sum_j\|f(j)\|<\infty\},$$ so it is a von Neumann algebra too.