Let us have a look on the proof of Theorem 2 in [P. G. Spain, [Boolean algebras of projections](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=3073512), *Proceedings of the Edinburgh Mathematical Society* (Series 2) **19**, 03, March 1975, 287-289]


The author claims in the proof of Theorem 2 that if $B$ is a (Bade) complete Boolean algebra of projections on a Banach space which is relatively weakly compact subset of the algebra of operators, then the weak (WOT?) closure of the algebra generated by $B$ is a W\*-algebra.


I don't think the proof is OK. Take $p\neq 2$, $p\in (1,\infty)$ and consider $X=\ell\_p$ with its canonical basis, which is 1-unconditional. The family $B$ of basis projections forms a complete Boolean algebra (actually isomorphic to $\wp(\omega)$) and is relatively weakly compact because all the projections have norm one and the unit ball of the space of operators on a reflexive space is WOT-compact. Though, $\overline{\mbox{alg }B}^{WOT}=B(\ell\_p)$ which is not a von Neumann algebra (it is not even Banach-space isomorphic to a von Neumann algebra).


My question is (before it's closed by the math police). Is Theorem 2 correct?



Is this result of Spain correct?
I haven't looked at the paper but your counterexample is mistaken. The basis projections generate not $B(l^p)$ but the algebra of multiplication operators, which is isometrically isomorphic to $l^\infty$ and hence is a von Neumann algebra.


