Answer to Question #287829 in Real Analysis for craciena

Question

 If S := {1/n - 1/m: n, mEN}, find inf S and sup S

Answer 1

$\text{Consider $n=1,2,\cdots,$ we get sequences \{1-$\frac{1}{m}$\}, $\{\frac{1}{2}$-$\frac{1}{n}$\}},\cdots. \text{We see that 1 is an upper}\\ \text{bound and claim that} \sup(S)=1. \text{ Now for any p$<1$, we have $\epsilon=1-p>0$. It is known}\\ \text{from the Archimedian property that there exists $m\in \mathbb{N} \ni \frac{1}{m}<1-p$, or equivalently,}\\ p<1-\frac{1}{m}. \text{therefore, any $p<1$ is not an upper bound of S.}\\ \text{Thus, this shows that $\sup(S)=1.$}\\ \text{Also, considering $m=1,2,\cdots,$we get sequences $,\{\frac{1}{n}-1\},\{\frac{1}{n}-\frac{1}{2}\},\cdots.$ It is clear that -1} \\ \text{is a lower bound of S since, for any $-1<q$, we get $q+1>0$ and thus there exists a}\\ \text{positive integer n$\in \mathbb{N} \ni \frac{1}{n}<q+1.$ It follows that we get $\frac{1}{n}-1\in S \ni \frac{1}{n}-1<q.$}\\ \text{Thus, this shows that $\inf(S)=-1.$}$