Prove that $\{ar+b:a,b\in Z\}$, where $r$ is an irrational number is dense in $R$ by using the following lemma

Question

Prove that $\{ar+b:a,b\in Z\}$, where $r$ is an irrational number is dense in $R$ by using the following lemma: If $x$ is irrational, there are infinitely many rational numbers $h/k$ with $k\gt 0$ such that $|x-h/k|\lt 1/k^2$.

My work:

Say $S=\{ar+b:a,b\in Z\}$. Given any real number $x$ and $\delta \gt 0$, by the above lemma, we can find some integer $k\gt 1/\delta$ such that $|kr-h|\lt \delta$.

Now, if we let $|kr-h|=L$, then $L\in S$, and I think since this number is less than $\delta$, we can multiply it by some integer $z$ to make it come inside the interval $(x-\delta, x+\delta)$. However, I cannot show this rigorously. How can I write this down? I would greatly appreciate any help.

Answer 1

Let $x\in\Bbb R$ and $\delta>0$. For every $\epsilon\in(0,\delta)$ there is some integer $n$ such that $x-\delta<n\epsilon<x+\delta$.

Proof: Using the Archimedean property, we see that there exists some $u\in\Bbb Z$ such that $u\epsilon\ge x+\delta$ and some $v_0\in\Bbb Z$ such that $v\epsilon\le x-\delta$ for any integer $v\le v_0$. Then the set $$A=\{m\in\Bbb Z: m\epsilon\ge x+\delta\}$$ is not empty and it is bounded below. Now we use the well ordering principle to see that $A$ has a minimum $t$, and define $n=t-1$. Then $n\notin A$, that is, $n\epsilon<x+\delta$. Also, $(n+1)\epsilon\ge x+\delta$, that is, $$n\epsilon\ge x+\delta-\epsilon>x-\epsilon>x-\delta$$

This completes the proof.