Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 1 / a / i / Solution

Put $s=\sup (-A)$. For every $x\in A$, $-x\le s$, hence $-s\le x$, so $-s$ is a lower bound of $A$. If $l$ is any lower bound of $A$, then $-x\le -l$ for every $x\in A$, making $-l$ an upper bound of $-A$. The definition of supremum gives $s\le -l$, or $l\le -s$. Thus $-s$ is the greatest lower bound, and