Ruzsa covering lemma

For finite subsets $S,T$ of an abelian group, with $T$ nonempty, suppose $|S+T|\le K|T|$. Then some $X\subseteq S$ satisfiesChoose a maximal family of pairwise disjoint translates $x+T$ with $x\in S$. Their union has size $|X||T|$ inside $S+T$, proving the cardinality bound. Maximality means that every $s+T$ meets some $x+T$. An equation $s+t=x+t^{\prime }$ gives $s\in x+T-T$, proving the covering. If $S$ is empty choose $X$ empty. An empty $T$ is excluded: it would make the assumed inequality vacuous while $X+T-T$ remained empty.