Question #110473
Use the order completeness property to show that the set S= {n/n+7:n belong to N} has a supremum and infimum
1
Expert's answer
2020-04-20T12:41:54-0400

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


i.e. S={18,29,310,...........}S=\{ \frac{1}{8},\frac{2}{9},\frac{3}{10},...........\}

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

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

Therefore 1 is an upper bound of SS .

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

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

We know that An upper bound uu of a non empty set SS in R\R is the Supremum of SS iff for every ϵ>0\epsilon>0 there is an xSx\in S such that

uϵ<xu-\epsilon <x .

As nn+7=17n+7\frac{n}{n+7}=1-\frac{7}{n+7} .So for every ϵ>0\epsilon >0 we find a nNn\in N such that


7n+7<ϵ\frac{7}{n+7}<\epsilon .

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

1ϵ<17n+7=nn+7=x.1-\epsilon <1-\frac{7}{n+7} =\frac{n}{n+7}=x.

Hence 1 is an least upper bound of S.

As 8nn+7, nN    nn+718, nN8n\geq n+7 , \forall \ n\in \N \implies \frac{n}{n+7}\geq \frac{1}{8} , \forall \ n\in \N .

Therefore 18\frac{1}{8} is a Lower bound of S.

Therefore , by Completeness property of R\R , SS has infimum .

As 18\frac{1}{8} is an element of S ,so by definition of infimum inf S=18.\text{inf} \ S =\frac{1}{8}.


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