# Prove that three following statements regarding the complete linear ordering of Cauchy sequences are equivalent

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

Let $$\mathcal{C}$$ be the set of Cauchy sequences of rationals. We define an equivalence relation $$\sim$$ on $$\mathcal{C}$$ by $$(a_n) \sim (b_n) \iff \forall \epsilon >0, \exists N, \forall n>N: |a_n - b_n| < \epsilon$$

Let $$\mathcal{C} / {\sim}$$ be the set of all equivalence classes of Cauchy sequences of rationals. We define a relation $$\preccurlyeq$$ on $$\mathcal{C} / {\sim}$$ by $$[(a_n)] \preccurlyeq [(b_n)] \iff \forall \epsilon >0, \exists N, \forall n>N: a_n - b_n < \epsilon$$

I have shown that $$\preccurlyeq$$ is a complete linear ordering here.

Theorem: Three following statemens are equivalent:

(1) $$[(a_n)] \prec [(b_n)]$$

(2) $$\exists \epsilon >0, \forall N, \exists n>N: \epsilon \le b_n -a_n$$

(3) $$\exists \epsilon >0, \exists N, \forall n>N: \epsilon \le b_n -a_n$$

My attempt:

(1) $$\implies$$ (2)

(1) $$\implies (a_n) \not \sim (b_n) \implies \exists \epsilon' > 0, \forall N, \exists n>N: |a_n - b_n| \ge \epsilon'$$.

Define $$f:\Bbb N \to \Bbb N$$ by $$f(N) = \min \{n \in \Bbb N \mid n > N \wedge |a_n - b_n| \ge \epsilon'\}$$

(1) $$\implies [(a_n)] \preccurlyeq [(b_n)] \implies \exists N', \forall n>N': a_n - b_n < \epsilon'$$.

Define $$g:\Bbb N \to \Bbb N$$ by

$$g(N)=\begin{cases}f(N') &\text{ if } N \le N'\\ f(N) &\text{ otherwise}\end{cases}$$

It follows that $$g(N) > N$$, $$|a_{g(N)} - b_{g(N)}| \ge \epsilon'$$, and $$a_{g(N)} - b_{g(N)} < \epsilon'$$. Hence $$a_{g(N)} - b_{g(N)} < -\epsilon'$$ or $$\epsilon' \le b_{g(N)} - a_{g(N)}$$ for all $$N \in \Bbb N$$.

Thus $$\exists \epsilon' >0, \forall N, \exists n>N: \epsilon' \le b_n -a_n$$.

As a result, (1) $$\implies$$ (2).

(2) $$\implies$$ (3)

(2) $$\implies \exists \epsilon' > 0, \forall N, \exists n>N: \epsilon' \le b_n -a_n$$.

$$(a_n),(b_n)$$ are Cauchy sequences $$\implies$$ $$\exists N', \forall m,n > N': |a_m - a_n| < \dfrac{\epsilon'}{3} \wedge |b_m - b_n|$$ $$< \dfrac{\epsilon'}{3}$$.

Let $$N'' =\min \{n \in \Bbb N \mid n > N' \wedge \epsilon' \le b_n -a_n\}$$. Then $$N'' > N'$$ and $$\epsilon' \le b_{N''} - a_{N''}$$.

$$n > N'' > N' \implies |a_n - a_{N''}| < \dfrac{\epsilon'}{3} \wedge |b_n - b_{N''}| < \dfrac{\epsilon'}{3}$$. This fact combining with $$\epsilon' \le b_{N''} - a_{N''}$$ implies $$\forall n> N'':a_n+\dfrac{\epsilon'}{3}.

As a result, (2) $$\implies$$ (3).

(3) $$\implies$$ (1)

(3) $$\implies$$ $$\exists \epsilon' >0, \exists N, \forall n>N: \epsilon' \le b_n -a_n$$ $$\implies$$ $$\exists \epsilon' >0, \exists N, \forall n>N: a_n - b_n \le -\epsilon' < 0$$ $$\implies$$ $$\forall \epsilon >0, \exists N, \forall n>N: a_n - b_n < \epsilon$$ $$\implies$$ $$[(a_n)] \preccurlyeq [(b_n)]$$.

(3) $$\implies$$ $$\exists \epsilon' >0, \exists N, \forall n>N: \epsilon' \le b_n -a_n$$ $$\implies$$ $$\exists \epsilon' >0, \exists N, \forall n>N: \epsilon' \le |b_n -a_n|$$ $$\implies$$ $$\exists \epsilon' >0, \exists N, \forall n>N: |a_n -b_n| \ge \epsilon'$$ $$\implies$$ $$(a_n) \not \sim (b_n)$$.

As a result, (3) $$\implies$$ (1).