Q1: Consider the statement of the Nested Intervals Property. As in the proof, consider the

set of all left ends . Show that every right end is an upper bound of .

Q2: Consider the set . Prove that .

A1: By the axiom of completeness, has a greatest upper bound . Thus, for , and for any there exists a such that . Therefore, for , is an upper buond for .

A2: Suppose there exists a lower bound such that . Then, , so we can consider . By lemma 1.3.7, there exists an such that, for any , . But, if we consider , then . Therefore, . This contradicts being a lower bound for . Hence, .

How do this proofs look? I feel I am not showing enough in the first one.

thanks