Let $(\mathfrak{M}_i, \mathcal{H})_i$ a family of Von neumann algebras acting on the same Hilbert space.
The von Neumann algebra
$$\bigcup_{i}\mathfrak{M}_i = \bigcap\{\mathfrak{M} \, : \, \mathfrak{M} \supset \mathfrak{M}_i \, \, \, \forall i\}$$
is called the generated von neumann algebra by $\mathfrak{M}_i$.
I need to prove that $(\bigcup_i \mathfrak{M}_i)' = \bigcap \mathfrak{M}_i'$ and $(\bigcap_i \mathfrak{M}_i)' = \bigcup_i \mathfrak{M}_i'$ (The commutants)
how may I do that?
Best Answer
Comment: the notation for the von Neumann algebra generated is terrible, as now you have no notation for the union. I will use the union as a union.
Notation: $W^*(S)$ is the von Neumann algebra generated by the set $S$. It is easy to check that $S'=W^*(S)'$: we have $S'=S'''=(S'')'=W^*(S)'$. So you actually only need to do your argument with the union, and not with the algebra generated.
Suppose that $T\in (\bigcup_j M_j)'$. Then, $T\in M_j'$ for any $j$, so $T\in\bigcap_jM_j'$. Conversely, if $T\in \bigcap_jM_j'$ then $T\in M_j'$ for all $j$; so $T\in(\bigcup_jM_j)'$.
Now, using the first equality. $$ W^*\Big(\bigcup_jM_j\Big)=\Big(\bigcup_jM_j\Big)'' =\Big[\big(\bigcup_jM_j\big)'\Big]'=\Big[\bigcap_jM_j'\Big]' $$
Edit: proof that the von Neumann algebra generated by a selfadjoint set $S\subset B(H)$ is $S''$.
Since $S''$ is a von Neumann algebra and it contains $S$, then $W^*(S)\subset S''$. Conversely, since $S\subset W^*(S)$, then $W^*(S)'\subset S'$, so $S''\subset W^*(S)''=W^*(S)$.