Assume, has a least upper bound .

By trichtonomy,

If,

Then, element of

And,

A contradiction.

If,

Then, is an upper bound.

And, .

A constradiction.

Thus, if the least upper bound exists.

Which is true because any negative is strictly less than a positive.

Use the completeness property. are upper bounded. Then, is upper bounded. Thus, by completeness it has a least upper bound.1) Let A and B be a subsets of the real numbers with least upper bound u and v. Prove that their union has a least upper bound, and express it in terms of u and v.