.... Then supK and infK both exist. Furthermore, supK∈K and infK∈K. Proof Recall that . Therefore, K has an and a . Because the and , we know K is......
...\in A\}.Thatis,L$ is the of s of A. Then infA (if it exists) is the of L. Remarks This is equivalent to definition 1, because s become s and maxim become minim in the......