Let $X:=\{(q_n)_n \subset \mathbb Q: (q_n)_n \mbox{ is Cauchy} \}$. Then, for $(q_n)_n, (r_n)_n\in X$, $(q_n)_n \sim (r_n)_n$ means that $(q_n)_n$ and $(r_n)_n$ are equivalent if for every $0<\varepsilon\in \mathbb{Q}$, there exists $N\in \mathbb{N}$ such that $n\ge N$ implies $|q_n-r_n|<\varepsilon$. Also, $(q_n)_n \le (r_n)_n$ if for every $0<\varepsilon \in\mathbb{Q}$, there exists $N\in\mathbb N$ such that $n\ge N$ implies $q_n\le r_n+\varepsilon$.
Now, suppose that $(q_n)_n \le (r_n)_n$, $(q_n)_n \sim (s_n)_n$, $(r_n)_n \sim (t_n)_n$. Prove that $(s_n)_n \le (t_n)_n$.
Now, I think the problem is probably easy to solve. But I'm having difficulty proving the above rigorously. What kind of approach should be used in this problem? Would appreciate some explanation.
Given $e>0.$ Take $N_1,N_2,N_3 \in \Bbb N$ such that $$(i) \quad n\geq N_1\implies q_n\leq r_n+e$$
$$(ii) \quad n\geq N_2\implies |q_n-s_n|\leq e$$
$$(iii) \quad n\geq N_3\implies |r_n-t_n|\leq e.$$ Let $N=\max (N_1,N_2,N_3).$ For $n\geq N$ we have $$s_n\leq q_n+e \quad \text {by (ii)}$$ $$\leq (r_n+e)+e \quad \text { by (i) }$$ $$\leq ((t_n+e)+e)+e \quad \text {by (iii) }$$ $$ =t_n+3e.$$ Now "reverse-engineer" this by replacing $e$ by $e/3$ (except in the first line above) and we have $$\forall n\geq N\;(s_n\leq t_n+e).$$ Remark: Regarding $N:$ The three inequalities $q_n\leq r_n+e, \;|q_n-s_n|\leq e,$ and $|r_n-t_n|\leq e$ each hold for all but finitely many $n,$ so all three of them hold for all but finitely many $n.$ So we can obtain a suitable $N$ this way instead.