= Solution
{c}
\b[True.] Write $\alpha=\sup A$ and $\beta=\inf B$. Every $a\in A$, $b\in B$ satisfies $a-b\leq\alpha-\beta$, giving an <upper bound> for the <set> $A-B$ of pairwise differences. For every $\varepsilon>0$, the definitions of <supremum> and <infimum> give $a\in A$ and $b\in B$ with
$$
a>\alpha-\varepsilon/2,\qquad b<\beta+\varepsilon/2.
$$
Then $a-b>\alpha-\beta-\varepsilon$. Hence no smaller <upper bound> works, and
$$
\boxed{\sup(A-B)=\sup A-\inf B.}
$$
Here $A-B$ means pairwise subtraction, rather than <set difference>.
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