Find with proof the infimum and supremum of a sequence

Question

We're given the sequence

$a_n=2-(1/n)$

And are asked to find both the infimum and supremum.

I understand the concept of both these terms, but putting them into practice is becoming a struggle.

Any suggestions?

Answer 1

Starting from $2-1/n<2$ for every natural number $n$ we see that $2$ is an upper bound of the set $A={\{a_n|n∈N\}}$.

We also see that for every $\epsilon>0$ there exists an $n ∈ N$ such that :

$2-\epsilon<2-1/n$ this implies that for every $\epsilon>0$ there exists an $n ∈ N$ such that $1/n<\epsilon$ , this is true from the Archimedean Property.

Therefore $\sup A=2$

To find the infimum of the set A we start from the inequality :

$1\leq n$ that is true for every $n ∈ N$

Then we get : $1/n\leq 1$ $\implies$ $1\leq 2-1/n$

So the set $A={\{a_n|n∈N\}}$ is bounded below from the number $1$ and $1$ ∈ A therefore we have :

$\inf A=\min A=1$.