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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.

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  • $\begingroup$ "This implies $\exists l \in A$ ... and $l = \inf A$"; no, the infimum of a set $A$ need not be a member of $A$. Also, "$\exists b \in B, \forall b \in B$..." what does this mean? $\endgroup$ – angryavian Sep 21 '17 at 2:55
  • $\begingroup$ i just realized that...would it be correct if i just said $\exists l \in \mathbb{R}$? and i believe i meant to write $\exists b \in A$, $\forall b \in B$ $\endgroup$ – rover2 Sep 21 '17 at 2:58
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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 $A$.

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  • $\begingroup$ how would i go about showing that sup $B$ is an element of $B$? $\endgroup$ – rover2 Sep 21 '17 at 3:13

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