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Let $R$ be a non-type-I factor acting on a separable Hilbert space.

Let $P(R)$ be the set of $R$'s projections with the usual ordering ($x \leq y \iff$ range$(x) \subseteq$ range$(y)$) under which it forms a complete lattice; let $P^+(R)$ be the same excluding the null projection.

Let $C$ be the poset of finite sequences of ordinals $< 2^\omega$, ordered by reverse inclusion (note $C$ is the most common poset used in forcing to collapse $2^\omega$ to $\omega$).

Question: Must there exist an order-embedding $\phi : C \rightarrow P^+(R)$ whose range is order-dense in the sense relevant to forcing (namely for all $x \in P^+(R)$ there exists $c \in C$ such that $\phi(c) \leq x$)?

I'm not sure how much intrinsically operator-algebraic interest this question holds, but it is potentially interesting for set theory, in particular for forcing. If there is such a embedding then $P^+(R)$ is forcing-equivalent to the standard continuum-collapse forcing; if not, I suspect $P^+(R)$ would (at least for some types of factor $R$) be inequivalent to known forcing posets, and would presumably have some novel properties.

Note that there definitely exist order-embeddings of $C$ into $P^+(R)$ that map each unbounded descending chain in $C$ to an unbounded descending chain in $P^+(R)$, but this property by itself doesn't ensure the embedding is dense.

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I believe I've shown (see answer to related stackoverflow question at https://math.stackexchange.com/a/4223869/250373 ) that there is such an embedding, on the supposition that every nontrivial lattice partition of $P(R)$ has continuum cardinality. This supposition is probably true in general but for now I can only prove this in the type III case, and with the assumption that CH holds.

[I'm deleting the sketch of a potential proof that had been in this answer previously, because it can't be salvaged.]

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