Compute $\sup(A)$ and $\inf(A)$ where $A= \{{1 + n^{-1} \mid n \in \mathbb{N}}\}$ 
Compute $\sup(A)$ and $\inf(A)$ where $A= \{{1 + n^{-1} \mid n \in \mathbb{N}}\}$


I know that $\frac{1}{n}$ approaches $0$, with large $n$ values, and so $1+ \frac{1}{n}$ approaches $1$, which is the inf.
The sup is $\infty$ ($0$ considered natural number) since $\frac{1}{n}$ approaches $\infty$.

Sup: $\infty$
Inf: $1$
Questions: 

How can I prove it? These bounding questions are really self-explanatory?

 A: I highly doubt $0\in\mathbb{N}$ is the convention used here. I think $\sup(A)$ is attained and should be $2$.
If you want to prove directly from the definition of infimum (also known as greatest lower bound) that $\inf(A)$ is $1$, you have to show that $1$ is a lower bound for $A$ and that it is the largest one.
$1$ is a lower bound for $A$ because $1\leq1+\frac{1}{n}$ for all $n\in\mathbb{N}$. Thus $1\leq a$ for every $a\in A$.
It is the largest one because for all $\epsilon>0$, $1+\epsilon>1+\frac{1}{N}$ if $N\in\mathbb{N}$ is chosen large enough. Indeed, this last inequality is equivalent to $\epsilon>\frac{1}{N}$ and the existence of such an $N\in\mathbb{N}$ is a consequence of the Archimedean property of the real numbers. Hence, any number $1+\epsilon$ greater than $1$ will not be a lower bound for $A$ since $a:=1+\frac{1}{N}\in A$ will be such that $1+\epsilon>a$.
A: We have $1 + \frac{1}{n} \geq 1 \; \forall n \in \mathbb{N}^*$. Thus $\inf A \geq 1$. 
Besides if  $1 + \frac{1}{n} \geq x \; \forall n \in \mathbb{N}^*$ then $x \leq 1$ using limits. Thus $\inf A \leq 1$. 
Finally $\inf A = 1$.
