Let $A$ be a nonempty and bounded below, and define $B= \{b\in \mathbb{R}: b$ is a lower bound for $A\}$. Show sup$B$ = inf$A$.
So far I have: Let $A$ be nonempty and bounded below. This implies $\exists l \in A, \forall a \in A, l\leq a$ and $l =$ inf$A$. Let $B = \{b\in \mathbb{R}: b$ is a lower bound for $A\}$ and let $M=$ sup$B$. This implies $\exists b\in B, \forall b \in B, b\leq M$. By definition of $B$, we know inf$A \in B$. Since $b\leq$ sup$B$, with sup$B$ the largest element in $B$, we have inf$A \leq M =$ sup$B$, so inf$A \leq$ sup$B$.
I realize I have to prove that sup$B \leq$ inf$A$ to get that sup$B$=inf$A$. I am having trouble doing so.
Part B: Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness.
Hint 1: Recalling that by definition, $\inf A$ is the largest lower bound for $A$, we know that $\inf A$ is the largest element of $B$. Thus it suffices to show that $\sup B$ is an element of $B$.
Hint 2: By definition of the supremum, we know that for any $\epsilon > 0$, there exists an element of $B$ in the interval $((\sup B) - \epsilon, \sup B]$. Use this fact to show that $\sup B$ is also a lower bound for $A$.