The given set is S={n+7n:n∈N} .
i.e. S={81,92,103,...........}
As n<n+7 , ∀ n∈N
⟹n+7n<1 ,∀ n∈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 : Supremum of S i,e 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 ϵ>0 there is an x∈S such that
u−ϵ<x .
As n+7n=1−n+77 .So for every ϵ>0 we find a n∈N such that
n+77<ϵ .
Hence , for every ϵ>0 there is an element of x∈S such that
1−ϵ<1−n+77=n+7n=x.
Hence 1 is an least upper bound of S.
As 8n≥n+7,∀ n∈N⟹n+7n≥81,∀ n∈N .
Therefore 81 is a Lower bound of S.
Therefore , by Completeness property of R , S has infimum .
As 81 is an element of S ,so by definition of infimum inf S=81.
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