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.