Let u = n, a = {x | x = 2K, K ∈ n}, B = {x | x = 2K + 1. K ∈ n}, find CUA, cub

Let u = n, a = {x | x = 2K, K ∈ n}, B = {x | x = 2K + 1. K ∈ n}, find CUA, cub

Because 2K + 1 is odd and 2K is even
So CUA = B, Cub = a