Question here about how the projections of a von Neumann algebra $\mathcal{R}$ might be arranged, relative to a projection that is not in $\mathcal{R}$.
Stipulate the following:
-- $H$ is a Hilbert space of countably infinite dimension;
-- $B(H)$ is the algebra of bounded linear operators on $H$;
-- $\mathcal{R} \subseteq B(H)$ is a von Neumann algebra without minimal nonzero projections;
-- $p \in B(H)$ is a projection NOT in $\mathcal{R}$, such that the identity operator is the only projection in $\mathcal{R}$ whose range contains $p$'s range;
-- we say that one projection "meets" another if the intersection of their ranges is not just $\{ 0 \}$.
Question: Can $p$ meet every nonzero projection in $\mathcal{R}$ yet commute with none of them (other than the identity)?
[Note that if $p$ as stipulated above commutes with all nonzero projections in $\mathcal{R}$, then it meets all of them. (For $p$ can now fail to meet a nonzero $q \in \mathcal{R}$ only by being orthogonal to it, but then $p$'s range would be contained by that of $(1 - q)$, which is a projection in $\mathcal{R}$ contradicting the stipulated uniqueness condition on the identity operator.) My question is how badly the converse to this can fail.]