TITLE: Suppose $b \in \mathbb{R}$ and $|b| < \frac{1}{n}$ for every positive integer n. Prove that $b = 0$.
QUESTION [6 upvotes]: This comes from an exercise in Appendix C from Axler's Measure, Integration & Real Analysis. The following is my approach. 
Suppose $b \neq 0$. Let $|b| = \epsilon$. Then 
        by Archimedean Property (2) 
        $$\exists n^* \in \mathbb{Z}^+ \text{ such that} \frac{1}{n^*} < \epsilon$$
        but $b < \frac{1}{n}$, $\forall n \in \mathbb{Z}^+$. Hence a contradiction.
Am I approaching this correctly, if I am are there any other approaches?

REPLY [2 votes]: Your approach is correct. There are several ways to define $\Bbb R$ from $\Bbb Q.$ They all result in isomorphic structures... Up to isomorphism there is only one ordered field in which every non-empty subset with a lower bound has a greatest lower  bound, which is why we say "the" real number system, not "a" real number system.
Suppose we only have $\Bbb Q$ and we hope to have an ordered field $F$ with the $glb$ property.  Then there cannot exist $b\in F$ such that $0\ne b$ and $|b|\leq 1/n$ for every $n\in \Bbb N.$ 
Proof: Suppose not. (i).  Consider the set $S=\{|b|/n:n\in \Bbb N\}.$ Then $|b|^2$ is a positive lower bound for $S$ because for every $n\in \Bbb N$ we have $|b|^2=|b|\cdot |b|\leq |b|\cdot (1/n).$
(ii). But $S$ has no $glb.$ Because if $c>0$ and $c$ is a lower bound for $S,$ then $c<2c.$ And for every $n$ $\in \Bbb N$ we have $c\leq |b|/2n,$ implying $2c\leq |b|/n.$ So $2c$ is also a lower bound for $S$.