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    #1
    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!

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    Euge's Avatar
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    #2
    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 $t < -x \le -\alpha$.

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