Given two type I (von-Neumann algebra) factors $\mathcal M,\mathcal N$, is there a *smallest* type I factor containing both $\mathcal M,\mathcal N$?


**Notes:**


* $\mathcal M,\mathcal N$ are over the same Hilbert space, of course.
* Obviously, a type I factor containing both exists (namely the set of all bounded operators). But I want the smallest. (I.e., it needs to be contained in all other type I factors containing both.)
* A standard approach would be to take the intersection of all type I factors containing both. However, it is not obvious that this is a type I factor. (The intersection of type I factors is not necessarily a type I factor, see the comments [here](https://mathoverflow.net/q/442854/101775).)
* Taking the von-Neumann algebra generated by $\mathcal M\cup\mathcal N$ (i.e., $(\mathcal M\cup\mathcal N)''$) also does not work, because that is not a type I factor in general (see the comments [here](https://mathoverflow.net/q/442906/101775)).
* If the answer would also cover the case where there are infinitely many factors (not just the two $\mathcal M,\mathcal N$) would be a bonus.
* The more elementary the proof of existence, the better. (I would need to formalize it in a computer-aided theorem prover, so the more "known facts" are used, the more of those I need to formalize first.)



Least upper bound of type I factors
No, there typically does not exist a smallest type I factor containing $\mathcal{M}$ and $\mathcal{N}$. Since the commutant of a type I factor is again a type I factor, by the bicommutant theorem, the question is equivalent with the existence of a maximal type I factor contained in $\mathcal{P} = \mathcal{M}' \cap \mathcal{N}'$. When no direct summand of $\mathcal{P}$ is of type I, such a maximal type I subfactor of $\mathcal{P}$ does not exist, because every projection can then be "halved".


For a specific counterexample, in the question, there is already a link to an argument showing that $\mathcal{P}$ could be a type II$\_1$ factor. Any type I subfactor of $\mathcal{P}$ must then be a matrix algebra. And any such matrix algebra is contained in a larger matrix subalgebra of $\mathcal{P}$.


