Well, when computing the a-successor for ({s0,s2},{s0}), we have to check the a-transitions from states s0 and s2. There is no transition with input a from s0, and from s2, the there is one to state s0. Hence, we get suc∃({s0,s2},a) = {s0} and suc∃({s0},a) = {}, the intersection with the accepting states does not change that, and thus we get state ({s0},{}).