1. Show that sup {1−1/n:nin N}=1{1−1/n:nin N}=frac{1}{2}. If S:={frac{1}{n}−frac{1}{m}:n,min N}S:={frac{1}{n}−frac{1}{m}:n,min N}, find inf S and sup S. end{tabular}

Wotzdorfg 2021-02-13 Answered
1. Show that sup {11n:nN}=1{11n:nN}=12. If S={1n1m:n,mN}S={1n1m:n,mN},fdfSandS.end{tabar}
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Expert Answer

Aniqa O'Neill
Answered 2021-02-14 Author has 100 answers

Let S=11n:nNSince11n<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 (1a)>11a>1/N011/N0>a.
This contradicts that aa is not an uper bound of SS. Hence 11 is the least upper bound of SS consequently
sup11/n:nN=1. (2) Let S=1n1/m:m,nN. Since 1n1m>1n1>11, 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>11N0<ϵ1N01<ϵ1=1+ϵ
Now since 1N01<1+ϵ for every ϵ>0 and 1N01S, thus
inf1n1/m:m,nN=1
Similarly we can prove that
sup1n1m:m,nN=1

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