Hi all, sorry if this is a dumb question, I don't know much about von Neumann algebras except the definition and a few relevant facts I've managed to prove by myself so I expect the answer will turn out to be well known. Anyway, let $\mathcal{H}$ be a Hilbert space, and suppose that $P$ is a commuting set of self-adjoint projections on $\mathcal{H}$, with the additional two properties:


1) $P$ is closed under complements, i.e. if $p \in P$ then so is $1 - p$.


2) $P$ is closed under suprema of arbitrary subsets, i.e. if $S \subseteq P$ then $\sup S \in P$ (here the projections on $\mathcal{H}$ are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$).


Now let $V$ denote the smallest von Neumann algebra containing $P$. Suppose that $p \in V$ is a self-adjoint projection. Is $p \in P$?


I know that $p$ is necessarily in the closure (relative to the weak operator topology) of the set of finite sums $\sum\_i \lambda\_i p\_i$, where $p\_i \in P$ and $\lambda\_i \in \mathbb{R}$. It seems like it may be possible to derive a contradiction from the assumption that $q$ has a strictly smaller range than $p$, where $q \equiv \sup ${$ r \in P | r \leq p $}. But I don't know how to proceed.



Question about von Neumann algebra generated by a complete algebra of projections
The answer "yes" follows from Theorem 2.8 of Bade's "[On Boolean algebras of projections and algebras of operators](http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0073954-0/home.html)," 1955, which is in the more general context of algebras of operators on a Banach space. 


Bade had previously proven a less general result that still covers your case, dealing with algebras of operators on reflexive spaces, in Theorem 3.4 of "[Weak and strong limits of spectral operators](http://projecteuclid.org/euclid.pjm/1103044796)," 1954.


