1. Show that sup {1−1n:n∈N}=1{1−1n:n∈N}=12. If S={1n−1m:n,m∈N}S={1n−1m:n,m∈N},f∈d∈fSand⊃S.end{taba―r}

Let $S=1-\frac{1}{n}:n\in NSince1-\frac{1}{n}<1$, 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
$sup1-1/n:n\in N=1.$ (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 $ϵ>0$, by Archimedean Property there exist a natural number N0 such that ​
$N0ϵ>1\Rightarrow \frac{1}{N0}<ϵ\Rightarrow \frac{1}{N0}-1<ϵ-1=-1+ϵ$
Now since $\frac{1}{N0}-1<-1+ϵ$ for every $ϵ>0and\frac{1}{N0}-1\in S$, thus
$inf\frac{1}{n}-1/m:m,n\in N=-1$
Similarly we can prove that
$sup\frac{1}{n}-\frac{1}{m}:m,n\in N=1$