For Q1, . But to show exists, you have to also show that is bounded above to invoke the least upper bound property. So I think you are using what you are trying to prove.
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
Let be your intervals for . They are nested, so, if . Assume, by contradiction, not every right end was an upper bound of . Thus, it means there exists, a left end so that which was not an upper bound for . That means, there exists so that . Now , therefore if then it must mean that . But we know that . Now but because . This is a contradiction by the nested interval property.
Can I simply break Q1 into cases?
Let . Since , we can see . Thus, there are three cases:
Case 1: If , then it is a given that .
Case 2: If , note that .
Case 3: If , then .
Therefore, is an upper bound for , for every .
Do I need to add any thing more to Q2?
Thanks for your help!