$(q_n)<(r_n)$ implies there is $q\in \mathbb{Q}$ such that $(q_n)<(q)<(r_n)$ Let $X$ be the set of Cauchy sequences of rational numbers.


*

*Then $(q_n)\sim (r_n)$ means that for every $0<\epsilon \in \mathbb{Q}$ there exists $N\in \mathbb{N}$ such that $|q_n-r_n|<\epsilon$ whenever $n\ge N$.

*$(q_n)\le (r_n)$ means that for every $0<\epsilon \in \mathbb{Q}$ there exists $N\in\mathbb{N}$ such that $q_n < r_n +\epsilon $ whenever $n\ge N$. 

*$(q_n) <(r_n)$ means that $(q_n)\le (r_n)$ but $(q_n)\not\sim (r_n)$.
Suppose that $(q_n)<(r_n)$ in $X$. Show that there exists $q\in\mathbb{Q}$ such that $(q)=(q,q,\dots)$ satisfies $(q_n)<(q)<(r_n)$.

My approach:

$(q_n)<(r_n)$ implies the following:


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*$\forall\epsilon>0,\exists N\in\mathbb{N}$ such that $n\ge N\implies q_n\le r_n+\epsilon$;

*$\exists \epsilon_0>0$ such that $\forall N\in\mathbb{N}, n\ge N \implies |q_n-r_n|\ge \epsilon_0$.
Let $\epsilon = \epsilon_0$, let $\epsilon_0 := q$. Then we have


*

*$q_n \le r_n + q$ (*)

*$-q\le q_n-r_n \le q$ (**)
But I'm stuck here and have no idea how to proceed. I've tried to manipulate the inequalities, but didn't achieve much success.
I'd appreciate if you could help me with this. Or is my approach not so good?
 A: Since $(q_n)\nsim (r_n)$, there exists $\epsilon_0>0$ such that for any $N$, there will always be some $n\geq N$ such that $|q_n - r_n| \geq \epsilon_0.$ 
Next, since $(q_n)$ and $(r_n)$ are Cauchy, there exist $N_q$ and $N_r$ such that $|q_n - q_{N_q}| <\epsilon_0/6$ for $n \geq N_q$ and $|r_n - r_{N_r}| <\epsilon_0/6$ for $n\geq N_r$. 
Next since $(q_n) \leq (r_n)$, there exists $N^*\geq \max\{N_q,N_r\}$ such that $q_n \leq r_n + \epsilon_0/6$ for all $n\geq N^*$.
Since $(q_n) \nsim (r_n)$ there is still some $m>N^*$ such that $|r_m - q_m| \geq \epsilon_0.$ 
Since $|r_m - q_m| \geq \epsilon_0$ and $q_m \leq r_m + \epsilon_0/6$, it follows that
$$r_m - q_m > \epsilon_0.$$ 
Further since $m>N^*$, $|q_n - q_m| \leq |q_n - q_{N_q}| + |q_{N_q} - q_m| \leq \epsilon_0/3 $, and $|r_n - r_m| \leq |r_n - r_{N_r}| + |r_{N_r} - q_n| \leq \epsilon_0/3 $ for all $n>m$.
Let $q = (q_{m} + r_m)/2$. Then $q\in \mathbb{Q}$ as the rationals are closed under addition and multiplication,and $q_n <  q_m +\epsilon_0/3 <  q < r_m - \epsilon_0/3 < r_n$ for all $n>m$.
