# Relation $(a_n)<(b_n)$ iff for every $\epsilon>0$, $a_n-b_n<\epsilon$ for all $n$ large enough, defines a total order $<$ on Cauchy sequences

Honestly, I figured out the idea to prove this theorem without much difficulty. But I found very hard to formalize it into a rigorous proof and it takes me three days to do it.

Could you please have a check on my attempt? Thank you so much!

A sequence $$\langle a_n \rangle_{n=0}^{\infty}$$ of rational numbers is a Cauchy sequence if for every rational $$\epsilon>0$$ there exists an integer $$N$$ such that $$|a_m - a_n| < \epsilon$$ for all $$m,n > N$$. Let $$\mathfrak{C}$$ be the set of all Cauchy sequences of rational numbers. We define a relation $$\preccurlyeq$$ on $$\mathfrak{C}$$ by $$\langle a_n \rangle_{n=0}^{\infty} \preccurlyeq \langle b_n \rangle_{n=0}^{\infty}$$ if and only if for every rational $$\epsilon>0$$ there exists an integer $$N$$ such that $$a_n - b_n < \epsilon$$ for all $$n > N$$.

Theorem: Any two elements of $$\mathfrak{C}$$ are comparable w.r.t $$\preccurlyeq$$.

My attempt:

Assume the contrary that there exist $$\langle a_n \rangle_{n=0}^{\infty}$$, $$\langle b_n \rangle_{n=0}^{\infty} \in \mathfrak{C}$$ that are incomparable. Then $$\langle a_n \rangle_{n=0}^{\infty} \not \preccurlyeq \langle b_n \rangle_{n=0}^{\infty}$$ and $$\langle b_n \rangle_{n=0}^{\infty} \not \preccurlyeq \langle a_n \rangle_{n=0}^{\infty}$$. As a result, $$\exists \epsilon >0,\forall N,\exists n>N:a_n - b_n \ge \epsilon$$ and $$\exists \epsilon >0,\forall N,\exists m>N:b_m - a_m \ge \epsilon$$.

Then $$\exists \epsilon >0,\forall N,\exists (n>N \wedge m>N):a_n - b_n \ge \epsilon \wedge b_m - a_m \ge \epsilon$$. By Axiom of Choice, there exist functions $$f:\Bbb N \to \Bbb N$$ and $$h:\Bbb N \to \Bbb N$$ such that $$f(N)>N$$, $$h(N)>N$$, $$a_{f(N)} - b_{f(N)} \ge \epsilon$$, and $$b_{h(N)} - a_{h(N)} \ge \epsilon$$.

Moreover, $$\langle b_n \rangle_{n=0}^{\infty}$$ is a Cauchy sequence $$\implies$$ $$\exists N_0,\forall m,n > N_0:|b_m - b_n| < \epsilon$$.

We define functions $$f':\Bbb N \to \Bbb N$$ and $$h':\Bbb N \to \Bbb N$$ by $$f'(N)=f(N_0)$$, $$h'(N)=h(N_0)$$ for all $$N and $$f'(N)=f(N)$$, $$h'(N)=h(N)$$ for all $$N \ge N_0$$.

It follows that $$f'(N)>N$$, $$h'(N)>N$$, $$a_{f'(N)} - b_{f'(N)} \ge \epsilon$$, $$b_{h'(N)} - a_{h'(N)} \ge \epsilon$$, and $$|b_{h'(N)} - b_{f'(N)}| < \epsilon$$ for all $$N \in \Bbb N$$. As a result,

\begin{align}|a_{f’(N)} - a_{h’(N)}| &= |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} ) + (b_{f’(N)} - b_{h’(N)})|\\ &\ge |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} )| - |b_{f’(N)} - b_{h’(N)}|\\ &= |a_{f’(N)} - b_{f’(N)}| + |b_{h’(N)} - a_{h’(N)} | - |b_{f’(N)} - b_{h’(N)}|\\ &> \epsilon + \epsilon - \epsilon =\epsilon\end{align}

Hence $$\exists \epsilon>0, \forall N, \exists(f’(N)>N \wedge h’(N)>N):a_{f’(N)} - a_{h’(N)}>\epsilon$$. This contradicts the fact that $$\langle a_n \rangle_{n=0}^{\infty}$$ is a Cauchy sequence.

I see only a (minor) problem with your attempt: at the first paragraph, those $$\varepsilon$$'s should be two distinct numbers a priori. But that's not a serious problem.
• Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $\preccurlyeq$ is NOT an order relation. This is because $\preccurlyeq$ is NOT antisymmetric, i.e there exist $\langle a_n \rangle_{n=0}^{\infty} \neq \langle b_n \rangle_{n=0}^{\infty}$ such that $\langle a_n \rangle_{n=0}^{\infty} \preccurlyeq \langle b_n \rangle_{n=0}^{\infty}$ and $\langle b_n \rangle_{n=0}^{\infty} \preccurlyeq \langle a_n \rangle_{n=0}^{\infty}$.[...] – Le Anh Dung Jan 30 at 11:03
• [...] 2. To avoid the use of Axiom of Choice, we can define $f:\Bbb N \to \Bbb N$ by $f(N)=\min \{n\in\Bbb N \mid n>N \wedge a_n - b_n \ge \epsilon \}$ for all $N\in\Bbb N$. – Le Anh Dung Jan 30 at 11:03