Let $\{P_i:i\in I\}$ be a set of projections in a von Neumann algebra $M$ and $J\subseteq I$ be finite. Then $P_J:= \bigvee_{i\in J} P_i$ is the smallest projection $P$ such that $\left(\sum_{i\in J}P_i\right)P=\left(\sum_{i\in J}P_i\right)$.
I'm able to show that $P_J$ is a projection and $\left(\sum_{i\in J}P_i\right)P_J=\left(\sum_{i\in J}P_i\right)$. But I'm unable to show it is the smallest.
If $Q$ is another projection such that $\left(\sum_{i\in J}P_i\right)Q=\left(\sum_{i\in J}P_i\right)$, then how to show $P_J\leq Q$. Since $P_J$ is the supremum, it suffices to show that each $P_i\leq Q$ but I couldn't show this either.
You can write the equality as $$ \Big(\sum_{i\in J}P_i\Big)(I-Q)=0. $$ Since the product of these selfadjoints is selfadjoint, they commute. From here we get $$ 0\leq (I-Q)P_j(I-Q)\leq \Big(\sum_{i\in J}P_i\Big)(I-Q)=0. $$ As $P_j$ is a projection, it follows that $(I-Q)P_j=0$, that is $QP_j=P_j$. From here we get directly that $QP_J=P_J$, as $Q$ acts as the identity on each of $P_jH$.