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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 $\forall n> N'':a_n+\dfrac{\epsilon'}{3}<b_n$.

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).

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