User:Lethe/supremum

proof that a poset X has least upper bound property iff it has the greatest lower bound property
Let L be the set of all lower bounds of the nonempty subset S of X. If L is nonempty, then any element of S is an upper bound of L, and so L has a supremum, let it be &lambda;.

Firstly, note that every upper and lower bound of any set S is comparable to every element of S (by definition). Thus the supremum and infimum of S, being members of the set of upper and lower bounds, respectively, are comparable to every element of S.

Thus the supremum of L is comparable to every element of L, and since every element of L is comparable to every element of S, and since comparability is transitive, the supremum of L is comparable to every element of S.

Now &forall;s&isin;S, &lambda; &le; s. For suppose not. Then &exist;s such that &lambda;&gt;s, since &lambda; is comparable to S. Then &forall;&#x2113;&isin;L, s &ge;&#x2113; because all &#8467; are lower bounds of S, and so s is a smaller upper bound of L than &lambda;, contradicting that &lambda; is a supremum, so we must have &lambda;&le;s. Thus &lambda; is a lower bound of S.

Finally, suppose &mu; is another lower bound of S. Then &mu;&isin;L and therefore &lambda;, being the supremum of L, is greater than &mu;.

So &lambda; is the infimum of S. If S has a lower bound, then it has a greatest lower bound.