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

set of all left ends $\displaystyle A =

\{a_{n} : n\in\mathbb{ N}\}$. Show that every right end $\displaystyle b_{n}$ is an upper bound of $\displaystyle A$.

Q2: Consider the set $\displaystyle A =

\{ \frac{1}{n} : n\in\mathbb{ N} \}$. Prove that $\displaystyle inf(A) = 0$.

A1: By the axiom of completeness, $\displaystyle A$ has a greatest upper bound $\displaystyle x=sup(A)$. Thus, for $\displaystyle \forall\\n\in\mathbb{N}$, $\displaystyle a_{n}\leq\\x$ and for any $\displaystyle \epsilon>0$ there exists a $\displaystyle b_{0}\in\\B$ such that $\displaystyle x<b_{0}<x+\epsilon$. Therefore, for $\displaystyle \forall\\n\in\mathbb{N}$, $\displaystyle b_{n}$ is an upper buond for $\displaystyle A$.

A2: Suppose there exists a lower bound $\displaystyle l$ such that $\displaystyle inf(A)=0<l$. Then, $\displaystyle l-0>0$, so we can consider $\displaystyle \epsilon=l-0>0$. By lemma 1.3.7, there exists an $\displaystyle a\in\\A$ such that, for any $\displaystyle \epsilon>0$, $\displaystyle a<l+\epsilon$. But, if we consider $\displaystyle \epsilon=l-0$, then $\displaystyle a<2l$. Therefore, $\displaystyle \frac{a}{2}<l$. This contradicts $\displaystyle l$ being a lower bound for $\displaystyle A$. Hence, $\displaystyle inf(A)=0$.

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

thanks