
MHB Apprentice
#1
September 12th, 2019,
10:10
Hey guys,
I'm kind of in a rush because I'll have to go to my classes soon here at USF Tampa, but I had one last problem for Intermediate Analysis that needs assistance. Thank you in advance to anyone providing it.
Question being asked: "Let $A$ be a nonempty set of real numbers which is bounded below. Let $A$ be the set of all numbers $x$, where $x$ is an element of $A$. Prove that $\inf A=\sup(A)$."
This is what I had so far for a proof: "Let $\alpha =\inf A$. For any $x\in A$, $\alpha \le x$. This implies that $\alpha \ge x$ for all $x\in A$. This means $\alpha$ is an upper bound of $A$. Also, if $\gamma <\alpha$ then $\gamma$ cannot be an upper bound of $A$ because if it is, then $\gamma \ge x$ for all $x \in A$. This implies that $\gamma >\alpha$ and that $\gamma$ is a lower bound of $A$, which is not possible since $\alpha = \inf A$."
Please help me out here, since this is the last problem I need help with at this point in time and I would really, really appreciate it. Thanks again in advance!

September 12th, 2019 10:10
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MHB Master
#2
September 14th, 2019,
15:53
Hi, AutGuy98!
You've correctly shown that $\alpha$ is an upper bound for $A$. To show that $\alpha = \sup(A)$, show that for every $t < \alpha$, there exists $x\in A$ such that $t < x \le \alpha$. Indeed, this will show that no number less than $\alpha$ is an upper bound for $A$, allowing you to conclude that $\alpha = \sup(A)$.
Let $t < \alpha$. Then $t > \alpha$; since $\alpha = \inf(A)$, there exists $x\in A$ such that $t > x \ge \alpha$. Thus $t < x \le \alpha$.