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Here are what I am allowed to use: Suppose $C$ denotes the set of all rational Cauchy sequences, and ~ denotes the relation on $C$, where $\{x_n\}$~$\{y_n\}$ means their term-wise difference tends to zero. Then the quotient $C$/~ denotes all equivalent classes of $C$ under ~.

Define the $>$ to be an order on $C$/~ such that if $x,y\in C/$~ and $y>x$, then there exists $\epsilon\in \mathbb{Q}^+$ such that for all $N>0$ there is some $n>N$ and $y_n\geq x_n+\epsilon$.

Show that for all $x,y\in C/$~, and $y>x$ there is some $r\in \mathbb{Q}$ such that $y>r>x$, note $r\mapsto [r]\in C/$~, where $[r]$ is the class of constant rational sequence.

Here is my approach.

Let $x,y\in C/$~ and $y>x$, and pick $\epsilon_1>0$ such that for all $N>0$ there is some $n_1>N$ and $y_{n_1}\geq x_{n_1} +\epsilon_1$. since $\{x_n\}\in x$ and $\{y_n\}\in y$ are cauchy, the above inequality holds for all $N$, pick $N=\max\{N_x,N_y\}$($N_x,N_y$ are the $N$s where $n,m>N_x\implies |x_n-x_m|<\epsilon$,... etc), then pick $N^\prime=n_{1}+1$ then there exists some $n_2\neq n_1$ and $y_{n_2}\geq x_{n_2}+\epsilon_1$, then $$x_{n_1}<x_{n_2}+\epsilon_1 \leq y_{n_2}<y_{n_1}+\epsilon_1$$

I'm thinking we can then pick $r=x_{n_2}+\epsilon_1$, since it's fixed, and then we have $y_{n_1}+\epsilon_1>r>x_{n_1}$. This is as far as I could get, I can't seem to get rid of the $\epsilon_1$.

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    $\begingroup$ I would start by showing that the definition of $>$ can be improved to something easier to work with here: $y>x$ if there is an $\epsilon>0$ and an $N$ such that $y_n>x_n+\epsilon$ for all $n>N.$ $\endgroup$ Commented Aug 30, 2022 at 2:35
  • $\begingroup$ @spaceisdarkgreen. thanks for the suggestions, i was able to move further with your idea, but still not finished, ill update and edit later $\endgroup$ Commented Aug 30, 2022 at 13:16
  • $\begingroup$ If you can show (or know) that the choice of sequence "upstairs" does not matter (it can't, since we've got an equivalence relation), then you can further simplify by assuming that your sequence $x_n$ is monotone increasing and $y_n$ is monotone decreasing. Then $y_n - x_n \geq epsilon > 0$. So you can fit a real interval in between the sequences, and that contains a rational $r$ $\endgroup$ Commented Sep 1, 2022 at 23:35

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So here is the solution I finally came up with, the suggestion in the comment did get me to a proof, but I then figured out a way without improving that definition.

Let $x,y\in \mathbf{C}/\sim$ and $x>y$. Then for $\{x_n\}\in x$ and $\{y_n\}\in y$, there exists $\epsilon_1\in\mathbb{Q}^+$, such that for all integer $N>0$ there exists $n_1>N$ and $x_{n_1}\geq y_{n_1}+\epsilon_1$. Since this holds for all $N>0$, pick $N_1$ large enough that it satisfies the Cauchy criterion for $\{x_n\}$ and $\{y_n\}$, and let $N_2\geq n_1+1$, then there exists $n_2>n_1>N_1$ such that $x_{n_2}\geq y_{n_2}+\epsilon_1$, and we have $$|x_{n_1}-x_{n_2}|<\frac{\epsilon_1}{4}\text{ and } |y_{n_1}-y_{n_2}|<\frac{\epsilon_1}{4}$$ Define $r=\frac{x_{n_1}+y_{n_1}}{2}$ which correspond to the constant cauchy sequence $\{r\}\in\mathbf{C}/\sim$, now we have \begin{align*}r+\frac{\epsilon_1}{4}<&\frac{x_{n_2}+y_{n_2}+\frac{\epsilon_1}{2}}{2}+\frac{\epsilon_1}{4} \\\leq&\frac{2x_{n_2}-\frac{\epsilon_1}{2}}{2}+\frac{\epsilon_1}{4}=x_{n_2} \end{align*} Hence we have $r+\frac{\epsilon_1}{4}<x_{n_2}\implies \{r\}<\{x_n\}$.

With similar ways, it can be shown there exists $\epsilon\in \mathbb{Q}^+$ such that for all $N>0$ there exists $n>N$ and $y_n+\epsilon\leq r$.

Completing the proof.

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