Answer to Question #110473 in Real Analysis for Pappu Kumar Gupta

Question

Use the order completeness property to show that the set S= {n/n+7:n belong to N} has a supremum and infimum

Answer 1

The given set is $S=\{ \frac{n}{n+7} : n\in \N \}$ .

i.e. $S=\{ \frac{1}{8},\frac{2}{9},\frac{3}{10},...........\}$

As $n<n+7 \ , \ \forall \ n\in \N$

$\implies \frac{n}{n+7} <1$ $,\forall \ n\in \N$ .

Therefore 1 is an upper bound of $S$ .

Again by completeness property of $\R$ ,every non empty subset of real number that has an upper bound also has a supremum in $\R$ .

Claim : $\text{Supremum of S i,e } \text{Sup}\ S=1$ .

We know that An upper bound $u$ of a non empty set $S$ in $\R$ is the Supremum of $S$ iff for every $\epsilon>0$ there is an $x\in S$ such that

$u-\epsilon <x$ .

As $\frac{n}{n+7}=1-\frac{7}{n+7}$ .So for every $\epsilon >0$ we find a $n\in N$ such that

$\frac{7}{n+7}<\epsilon$ .

Hence , for every $\epsilon>0$ there is an element of $x\in S$ such that

$1-\epsilon <1-\frac{7}{n+7} =\frac{n}{n+7}=x.$

Hence 1 is an least upper bound of S.

As $8n\geq n+7 , \forall \ n\in \N \implies \frac{n}{n+7}\geq \frac{1}{8} , \forall \ n\in \N$ .

Therefore $\frac{1}{8}$ is a Lower bound of S.

Therefore , by Completeness property of $\R$ , $S$ has infimum .

As $\frac{1}{8}$ is an element of S ,so by definition of infimum $\text{inf} \ S =\frac{1}{8}.$