Take the set B, which is bounded above, since for b \in B, and a \in A, a > b, so all a are upperbounds of b. Call sup B = c. Then for any d< c, d is not an upperbound of B, so d is not in A. So c <= a \in A, and c \in B. Then for any e st e>c, then e is not in B, since c is the supremum of B. Then c is also the inf A, so inf A = sup B