 Wotzdorfg

2021-02-13

1. Show that sup $\left\{1-\frac{1}{n}:n\in N\right\}=1\left\{1-\frac{1}{n}:n\in N\right\}=\frac{1}{2}$. If $S\phantom{\rule{0.222em}{0ex}}=\left\{\frac{1}{n}-\frac{1}{m}:n,m\in N\right\}S\phantom{\rule{0.222em}{0ex}}=\left\{\frac{1}{n}-\frac{1}{m}:n,m\in N\right\},f\in d\in fS\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\supset S.end\left\{tab\underset{―}{a}r\right\}$ Aniqa O'Neill

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 $\left(1-a\right)>1⇒1-a>1/N0⇒1-1/N0>a$.
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 ​
$N0ϵ>1⇒\frac{1}{N0}<ϵ⇒\frac{1}{N0}-1<ϵ-1=-1+ϵ$
Now since $\frac{1}{N0}-1<-1+ϵ$ for every , thus
$inf\frac{1}{n}-1/m:m,n\in N=-1$
Similarly we can prove that
$sup\frac{1}{n}-\frac{1}{m}:m,n\in N=1$

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