$\inf$ and $\sup$ of the following sets (p12 Exercise 1.3.6 Understanding Analysis, Abbott) I try to determine the $\inf$ and $\sup$ of the following sets. Can you tell me if my result is correct? I am also interested in how to compute it. Now I only determined it by varying the values. 
Define sets $B= \{ {n \over n + m } : m,n \in \mathbb N\}$, $C = \{ {n \over 2n + 1} :n \in \mathbb N\} $, $D= \{ {n \over m} : n , m \in \mathbb N , n+m \le 10 \}$. The $\inf$ and $\sup$ is:
$\inf B =0, \sup B = 1$ , 
$\inf C = 0, \sup C = {1 \over 2}$, 
$\inf D = 0, \sup D = \infty$.
 A: To approach such problems I would suggest first to let one index (or both) run to infinity and take a look at the limit, should it exist.
As for set $B$,
$$
\lim_{m\to\infty}\frac{n}{m+n} ~=~ 0 ~~\forall n
\quad\text{and}\quad
\lim_{n\to\infty}\frac{n}{m+n} ~=~ 1 ~~\forall m
$$
Since $m,n$ are are positive, so is $\frac{n}{m+n}$. The fact that you find a sequence of points that converges to $0$ (fix, for instance, $n=1$ and consider the sequence $\{a_m=\frac{1}{1+m}\}_{m\in\mathbb N}\subset B$, by definition you have that you approach as much as you want to $0$, and you can indeed conclude that $\inf B =0$.
From the second limit you should observe that, since $n\leq m+n$, then $\frac{n}{m+n}\leq 1$, and apply a similar reasoning to deduce that $\sup B=1$.
I'll leave you the computations for set $C$, which are even easier. As @Oleg567 points out regarding $\inf C$, it would be helpful to observe that the sequence increases wrt variable $n$, so that it achieves its infimum (minimum, then) for the least value of $n$, i.e. $n=1$.
As @Quickbeam2k1 commented, the constraint $m+n\leq 10$ makes set $D$ finite, so no limits should be evaluated. It's pretty much a matter of listing all the values, or, better, to use the fact that $\frac nm$ increases wrt $n$ and decreases wrt $m$.
A: Hint: Recall the following version of  Archimedean Property of $\mathbb{R}$:

$$\forall \epsilon>0\ \exists n\in \mathbb{N}:\frac1n<\epsilon$$

