Question

Determine, with proof, the supremum and infimum of the set A={(n+1)/n | n ∈ N}

Answer 1

Consider the set $A=\{\dfrac{n+1}{n}:n\in\N\}$

Since $\dfrac{n+1}{n}=1+\dfrac{1}{n}>1$ for each $n\in\N, 1$ is a lower bound of the set. Moreover, for each $\varepsilon>0,$ by Archimedean Property, there exists $N\in\N$ such that $N\geq\dfrac{1}{\varepsilon}.$ Then $\dfrac{1}{N}\leq\varepsilon$ and therefore $1+\varepsilon$ is not a lower bound of the set for each $\varepsilon>0.$ We conclude that $1$ is the infinum of the set $\{\dfrac{n+1}{n}:n\in\N\}.$

Since $n\geq1$ we have $\dfrac{n+1}{n}=1+\dfrac{1}{n}\leq1+\dfrac{1}{1}=2, n\in\N$

The set $A$ is bounded above: there is $M=2$ such that $\dfrac{n+1}{n}\leq2$ for all $n\in\N.$

The number $M=2$ is the upper bound of the set $A.$

Since $2\in A$ then we conclude that $2$ is the supremum of the set $\{\dfrac{n+1}{n}:n\in\N\}.$