Find the supremum and infimum of $B:=\{\frac{1}{n}:n\in \mathbb{N}\}$ without using limits If I am supposed to find the infimum and supremum of $B:=\{\frac{1}{n}:n\in \mathbb{N}\}$ without using limits is it sufficient to use the fact that $\forall n\in N\Rightarrow n\geq 1\Rightarrow \frac{1}{n}\leq 1$ to show that the supremum is 1 and then that $\frac{1}{n}>0$ to show that the infimum is $0$. 
 A: It isn't quite sufficient. That is enough to show that $1$ is an upper bound of $B$ (to show it is the least such, just show that $1\in B$) and that $0$ is a lower bound of $B$ (to show that it is the greatest such, note that if $x>0,$ then there is some $n\in\Bbb N$ such that $n>\frac1x,$ and go from there).
A: Note that an upper bound for a set need not be the least upper bound for that set; so what you want to do is not sufficient. 
We claim that $\sup B = 1$; let $\varepsilon > 0$. If $\varepsilon \geq 1$, then $n := 1$ is such that $1 - \varepsilon < 1/n \leq 1$; if $\varepsilon < 1$, then $n := \lfloor \frac{1}{1-\varepsilon} \rfloor$ is such that $1 - \varepsilon < 1/n \leq 1$; hence there is always some $n \geq 1$ such that $1- \varepsilon < 1/n \leq 1$, 
and we are done.
In the similar fashion, we have $\inf B = 0$; for $n := \lceil 1/\varepsilon \rceil + 1$ is such that $0 < 1/n < 0 + \varepsilon = \varepsilon$.
A: You also need to show those are the least and greatest upper and lower bounds respectively. Meaning if $l$ is a greater lower bound, then there is something in the set less than $l$ and same for the other. 
Hint: use a consequence of the Archimedian principle for the inf. The sup is not difficult.
