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

The infimum of $A$ is its greatest lower bound: $\alpha =\inf A$ means that $\alpha \le x$ for every $x\in A$, and that every lower bound $l$ of $A$ satisfies $l\le \alpha$. The supremum is its least upper bound: $\beta =\sup A$ means that $x\le \beta$ for every $x\in A$, and every upper bound $u$ satisfies $\beta \le u$. The completeness of the real numbers ensures that both exist for a nonempty bounded set of real numbers.

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

Let $M=\max \{\sup A,\sup B\}$. Every element of the union $A\cup B$ is at most $M$, so $M$ is an upper bound. Conversely, every upper bound of $A\cup B$ is an upper bound of each of $A$ and $B$, hence is at least both suprema and therefore at least $M$. Consequently