Question

Use the order completeness property to show that the set

S={n/(n+1): n∈N}


has a

supremum and infimum.

Answer 1

1 STEP: Search for the supremum of a given sequence

$$\frac{n}{n+1}=\frac{(n+1)-1}{n+1}=1-\frac{1}{n+1}$$

Conclusion,

$$S_n=1-\frac{1}{n+1}<1,\quad\forall n\in\mathbb{N}\\[0.5cm] \boxed{\sup(S_n)=1}$$

2 STEP: Search for the infimum of a given sequence

$$1+n\le n+n,\forall n\in\mathbb{N}\longrightarrow\\[0.5cm] \frac{1}{1+n}\ge\frac{1}{n+n}\longrightarrow\frac{n}{n+1}\ge\frac{n}{n+n}=\frac{1}{2}$$

Conclusion,

$$S_n\ge\frac{1}{2},\quad\forall n\in\mathbb{N}\\[0.5cm] \boxed{\inf(S_n)=\frac{1}{2}}$$