Let \(S={1−\frac{1}{n}:n\in N} Since 1−\frac{1}{n}<1\), therefore 1 is an upper bound of S. We have to prove that 1 is the least upper bound. Let aa be another upper bound of SS such that a<1. Since 1−a>01−a>0 by Archimedean Property there exist a natural number N0 such that

N0 \((1−a)>1 \Rightarrow 1-a>1/N0 \Rightarrow 1-1/N0>a\).

This contradicts that aa is not an uper bound of SS. Hence 11 is the least upper bound of SS consequently

\(\sup{1-1/n:n\in N}=1.\) (2) Let \(S={\frac{1}{n} −1/m:m,n\in N}\). Since \(\frac{1}{n}-\frac{1}{m}>\frac{1}{n}-1>1-1\), therefore −1 is a lower bound of S. We have to prove that 1 is the gratest upper bound. Let \(\epsilon >0\), by Archimedean Property there exist a natural number N0 such that

\(N0\epsilon >1 \Rightarrow \frac{1}{N0}<\epsilon \Rightarrow \frac{1}{N0}-1<\epsilon -1=-1+\epsilon\)

Now since \(\frac{1}{N0}−1<−1+\epsilon\) for every \(\epsilon >0 \ and \ \frac{1}{N0}-1\in S\), thus

\(inf{\frac{1}{n}-1/m:m,n\in N}=-1\)

Similarly we can prove that

\(\sup{\frac{1}{n}-\frac{1}{m}:m,n\in N}=1\)