F be an ordered field. Suppose that every non-empty subset of F that is bounded below has an glb (inf). We show that every non-empty subset of F tat is bounded above has a lub (sup). So suppose that A is bounded above, say by b. Let ?A be the set of all ?x, where x?A. We show that ?A is bounded below. For let ?x??A. Since x?b, we have ?b??x, and therefore ?b is a lower bound of ?A. By the glb property, ?A has a glb, which it is convenient to call ?c. We show that c is a lub of A. (Note: we are saying \"a\" glb, but we know that glb, and lub are unique when they exist.) So first we show that c is an upper bound of A, that is, that x?c for all x?A. We know ?c is a lower bound of ?a. So ?c??x for all x?A. It follows that x?c for all x?A Solution F be an ordered field. Suppose that every non-empty subset of F that is bounded below has an glb (inf). We show that every non-empty subset of F tat is bounded above has a lub (sup). So suppose that A is bounded above, say by b. Let ?A be the set of all ?x, where x?A. We show that ?A is bounded below. For let ?x??A. Since x?b, we have ?b??x, and therefore ?b is a lower bound of ?A. By the glb property, ?A has a glb, which it is convenient to call ?c. We show that c is a lub of A. (Note: we are saying \"a\" glb, but we know that glb, and lub are unique when they exist.) So first we show that c is an upper bound of A, that is, that x?c for all x?A. We know ?c is a lower bound of ?a. So ?c??x for all x?A. It follows that x?c for all x?A.