I am stuck on this problem for a while now, Given set $S\subset \mathbb{R}$ not empty and bounded above, let $x_n=\min\{\frac{m}{n}: m\in \mathbb{Z} $ and $\frac{m}{n} $ is an upper bound for $ S\}$. Ultimately, I'm trying to show $\mathbb{R}$ satisfies the least upper bound property, using only the idea that $\mathbb{R}$ is a collection of an equivalent class of rational Cauchy Sequence with the equivalence relation where two sequences are related if their termwise difference tends to zero. So the first thing I need to do is to show this sequence is indeed Cauchy,
Here is my scratch work on trying to find an big N, suppose $n,m >N$ then $|x_n-x_m|=|\frac{k_n}{n}-\frac{k_m}{m}|<|\frac{k_nm-k_mn}{nm}|<|\frac{k_nm-k_mn}{N^2}|$, But now I can't get rid of the variables on the top with some fixed bounds. I gave it long and hard thoughts but I seem to not be able to come up with a way to bound the top part. Can someone give me some hints?
Hint 1 Between $\frac{k}{n}$ and $\frac{k+1}{n}$, there is an element of the form $\frac{p}{m}$ if $m>n$.
Hint 2 To show that a sequence is Cauchy, try to find a sequence $a_n$ tending to 0 such that $|x_n-x_m|\le a_n$ when $m > n$.