- #1

AutGuy98

- 20

- 0

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!