TITLE: Find the supremum and infimum of $B:=\{\frac{1}{n}:n\in \mathbb{N}\}$ without using limits
QUESTION [1 upvotes]: 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 $0$.

REPLY [1 votes]: 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 $0 < 1/n < 0 + \varepsilon = \varepsilon$.
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