How to prove that infimum of this set is $0$

Question

How do I prove that the infimum of the set $A=\{x+\frac{1}{n}:x\in(0,1),n\in\mathbb{Z^+}\}$ is $0$?

Clearly $0$ is a lower bound of A since $1/n$ is greater than $0$ for all $n$. How do I show that any real number $K>0$ is not also a lower bound, hence making $0$ the infimum?

Thanks!

Answer 1

Use the fact that $\frac K2+\frac1n\lt K$ if $n$ is large enough.

Answer 2

It remains to show that for a given $\epsilon>0$ there are $x\in(0,1)$ and $n\in\mathbb{Z^+}$ such that $$ x+\frac{1}{n}<\epsilon.$$ Note that the above inequality is implied by $x<\frac{\epsilon}{2}$ and $\frac{1}{n}<\frac{\epsilon}{2}$. Can you take it from here?