Let $\mathcal{R}$ be a type III factor acting on separable Hilbert space $H$, and $S \subseteq H$ a closed linear subspace such that every nonzero $v \in S$ is cyclic for $R$; can there exist a nonzero projection $P \in \mathcal{R}$ whose range is disjoint from $S^\perp$ (aside from the null vector)?
(A vector $v$ is "cyclic for $\mathcal{R}$" if its range under the action of $\mathcal{R}$ is dense in $H$, i.e. $[\mathcal{R}v] = H$.)
We may assume $S$ is infinite-dimensional since if it has finite dimension the answer is "no": $S^\perp$ would then have finite co-dimension, and the range of every nonzero $P \in \mathcal{R}$ has infinite dimension since $\mathcal{R}$ is type III, and two subspaces always have nontrivial intersection if one's dimension is greater than the other's co-dimension.
Some observations:
-- There do exist infinite-dimensional closed subspaces whose vectors are all cyclic for $\mathcal{R}$;
-- But when $v, w$ are cyclic for $\mathcal{R}$, it is not always the case that all linear combinations of $v$ and $w$ are cyclic for $\mathcal{R}$.
-- Presumably we need to find a condition, in terms of projections in $\mathcal{R}$ and/or its commutant, that holds for a collection of cyclic-for-$\mathcal{R}$ vectors just when its closed span consists entirely of cyclic-for-$\mathcal{R}$ vectors. But I haven't been able to make much progress towards finding such a condition.