# Relation of Supremum and Infimum [duplicate]

Let $\mathcal{S}$ be a non-empty bounded set of real numbers, and define for a real number $a$ the set $aS=\left\{ax|\ \ x \in\mathcal{S}\right\}$. Prove that $\sup(aS)=a \inf S$ and $\inf(aS)=a \sup S$ if $a<0$.

## marked as duplicate by Git Gud, Namaste, njguliyev, Andrey Rekalo, user61527 Sep 7 '13 at 19:03

Hint. Write $\mathrm{ub}(X)$ for the upper bounds of $X$ and $\mathrm{lb}(X)$ for the lower bounds. Then we define
$$\sup(X) = \min \mathrm{ub}(X),\quad \inf(X) = \max \mathrm{lb}(X).$$
Thus it is sufficient to show that for $a < 0$ we have:
1. $\mathrm{ub}(aX)=a\,\mathrm{lb}(X)$.
2. $\mathrm{min}(aX)=a\,\mathrm{max}(X)$.