# Proving infinums and supremums sets of real numbers

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

$$\inf A = -\sup(-A).$$

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What have you tried? –  Nate Eldredge Sep 28 '11 at 23:35

Prove something more general: the set of lower bounds of $A$ is the negative of the set of upper bounds of $-A$. In symbols, using a notation that I hope is clear: $\mathcal{L}(A) = -\mathcal{U}(-A)$. Consider that by definition $\inf X=\max \mathcal{L}(X)$ and $\sup X=\min\mathcal{U}(X)$; and clearly $\max(Y) = -\min (-Y)$.

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Let $x=\inf A$. What do you know about $x$? You know that $x\le a$ for each $a\in A$, and you know that if $y$ is such that $y\le a$ for each $a\in A$, then $y\le x$. You want to prove that $x=-\sup(-A)$, or, equivalently, that $-x=\sup(-A)$.

1. What properties does $\sup(-A)$ have to have? Use the definition of the supremum.
2. How can you use the known properties of $x$ to show that $-x$ has the properties that you listed for (1)?
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HINT $\rm\ x < \inf\: A \iff x< A \iff\!\! -x > -A \iff\!\! -x > \sup(-A)\iff x < -\sup(-A)$

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