Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 1 / a

True. Put $\alpha =\sup A$, $\beta =\sup B$ and $u=\max \{\alpha ,\beta \}$. These suprema exist by completeness of the real numbers, since the sets are nonempty and bounded. Every set element of the union $A\cup B$ is at most $u$, so $u$ is an upper bound. Conversely, every upper bound of $A\cup B$ bounds both $A$ and $B$, and is therefore at least both suprema. Thus

True. Write $\alpha =\sup A$ and $\beta =\inf B$. Every $a\in A$, $b\in B$ satisfies $a-b\le \alpha -\beta$, giving an upper bound for the set $A-B$ of pairwise differences. For every $\epsilon >0$, the definitions of supremum and infimum give $a\in A$ and $b\in B$ withThen $a-b>\alpha -\beta -\epsilon$. Hence no smaller upper bound works, andHere $A-B$ means pairwise subtraction, rather than set difference.

False. Take the finite sets $A=B=\{-2,-1\}$. Their suprema are both $-1$, whereas the set of pairwise products is $A\cdot B=\{1,2,4\}$. ConsequentlyNegative factors can reverse an inequality, which prevents the proposed argument with separate suprema.