If $\mathcal{R}$ is a von Neumann factor acting on a separable Hilbert space $H$ and $P_1 < P_2 < P_3 < \cdots$ is a strictly increasing sequence of projections in $\mathcal{R}$ whose only upper bound is the identity operator, must there exist a nonzero projection $Q \in \mathcal{R}$ such that $Q \wedge P_n = 0$ for all $n$?
If $\mathcal{R} = B(H)$ (the algebra of all bounded operators on $H$) then clearly yes: for each $n$ let $v_n$ be a nonzero vector in the range of $(P_{n+1} - P_n)$, and let $v \in H$ be some linear combination of all the $v_n$ (with all coefficients nonzero); then $v$ is not in the range of any $P_n$, and the one-dimensional projection along $v$ serves as the required $Q$.
But must there exist such a $Q \in \mathcal{R}$ when $\mathcal{R} \neq B(H)$?
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ADDENDUM: I can at least sketch a proof that the answer is yes when $R$ is type III; I'd appreciate any comments although it is a bit long.
It depends on this claim which I believe is right although I won't swear to it: When $\mathcal{R}$ is a factor and $U \in \mathcal{R}$ is unitary and maps closed subspace $A$ onto an orthogonal subspace $B$, then for all $\theta$ in $(0,\pi / 2)$ there exists a unitary $U_\theta \in \mathcal{R}$ that "rotates partway along $U$" such that for all vectors $v \in A$, $U_\theta v = v \cos \theta + (U v) \sin \theta$. Thus if $P \in \mathcal{R}$ is the projection onto $A$, the operator $U_{\theta} P$ is a partial isometry whose initial space is $A$, and whose final space is a subspace of $A \vee B$ that is disjoint from both $A$ and $B$ (other than the null vector), and for all unit-norm $v \in A$, $||v - (U_{\theta} P) v|| \leq \sqrt{2} \sin(\theta)$.
Now when $\mathcal{R}$ is type III, all nonzero projections $P, Q \in \mathcal{R}$ are Murray-von-Neumann-equivalent, i.e. there exists a partial isometry $V \in \mathcal{R}$ whose initial space is $P$'s range and whose final space is $Q$'s range. Unless $P$ or $Q$ is the identity, $V$ can also be extended to a unitary $U \in \mathcal{R}$, because there also exists a partial isometry $V' \in \mathcal{R}$ mapping the range of $P^\perp$ onto the range of $Q^\perp$, so $V + V'$ will be such a $U$.
Then, given the sequence $P_n$ as above, let $T_0$ be the projection $P_0$, and inductively define $T_{n+1}$ as $U_{1/{n^2}} T_n$ (using the $U_\theta$ notation from above) where $U$ is a unitary operator extending a partial isometry $V$ from the range of $T_n$ to the range of $(P_{n+1}-P_n)$.
Claim: the $T_n$ converge in norm to an operator $T$, and the nonzero projection $Q$ onto $T$'s range exists in $\mathcal{R}$, and $Q \wedge P_n = 0$ for all $n$.
Proof: The $T_n$ form a cauchy sequence under the operator norm by the following argument. An arbitrary unit-norm $v \in H$ can be written uniquely as $v_1 + v_2$ with $v_1$ in $P_0$'s kernel and $v_2$ in $P_0$'s range (and both $v_1$ and $v_2$ have norm $\leq 1$). By the way we defined the $T_n$, we have $T_n v_1 = 0$ for all $n$, and $|| T_{n+1} v_2 - T_n v_2 || \leq \sqrt{2} \sin (1/{n^2})$. Thus $|| T_{n+1}v - T_n v || \leq \sqrt{2} \sin (1/{n^2})$, and since $v$ was arbitrary, $|| T_{n+1} - T_n || \leq \sqrt{2} \sin (1/{n^2}) \leq \sqrt{2}/{n^2}$.
So the $T_n$ form a cauchy sequence under the operator norm, and since $B(H)$ is a complete banach space under that norm, and $\mathcal{R}$ (being a C* algebra) is norm-closed, the limit operator $T$ exists in $\mathcal{R}$. By polar decomposition (possible in all von Neumann algebras) the projection $Q$ onto $T$'s range is also in $\mathcal{R}$.
Finally we give an informal geometric argument that any nonzero $w$ in $T$'s range is not in the range of any $P_n$. By construction, for $n>0$, $T_n$'s range is contained in $P_n$'s and is disjoint (aside from the null vector) from $P_{n-1}$'s. So if $v$ is any nonzero vector in $P_0$'s range, $T_n$ maps it outside $P_{n-1}$'s range, and each successive $U_{1/{n^2}}$ will "rotate it towards" a subspace orthogonal to $P_{n-1}$'s range, which cannot move it closer to $P_{n-1}$'s range. Thus the vector toward which $T_i v$ converges as $i \rightarrow \infty$ cannot be in $P_{n-1}$'s range.