# Prove that $\preccurlyeq$ is a linear ordering over the set of all equivalence classes of Cauchy sequences of rationals

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 $$[(a_n)] \preccurlyeq [(b_n)] \iff \forall \epsilon >0, \exists N, \forall n>N: a_n - b_n < \epsilon$$

Theorem: $$\preccurlyeq$$ is a linear ordering.

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

My attempt:

1. $$\preccurlyeq$$ is well-defined

It suffices to prove $$(a_n) \sim (x_n)$$ and $$(b_n) \sim (y_n)$$ and $$[(a_n)] \preccurlyeq [(b_n)]$$ $$\implies$$ $$[(x_n)] \preccurlyeq [(y_n)]$$.

First, $$(a_n) \sim (x_n) \implies \exists N_1, \forall n>N_1: |a_n - x_n| < \dfrac{\epsilon}{3} \implies \exists N_1, \forall n>N_1:$$ $$x_n - a_n < \dfrac{\epsilon}{3}$$. Second, $$(b_n) \sim (y_n) \implies \exists N_2, \forall n>N_2: |b_n - y_n| < \dfrac{\epsilon}{3} \implies$$ $$\exists N_2, \forall n>N_2: b_n - y_n < \dfrac{\epsilon}{3}$$. Finally, $$[(a_n)] \preccurlyeq [(b_n)] \implies \exists N_3, \forall n>N_3:$$ $$a_n - b_n < \dfrac{\epsilon}{3}$$.

Let $$N= \max \{N_1,N_2,N_3\}$$. It follows that

$$\forall \epsilon>0, \exists N, \forall n> N: x_n - y_n = (x_n - a_n) + (b_n-y_n) +(a_n-b_n) < \dfrac{\epsilon}{3} +\dfrac{\epsilon}{3}$$ $$+\dfrac{\epsilon}{3} = \epsilon$$. Hence $$[(x_n)] \preccurlyeq [(y_n)]$$.

1. $$\preccurlyeq$$ is reflexive

$$\forall \epsilon >0, \exists N, \forall n>N: a_n - a_n = 0 < \epsilon$$ $$\implies$$ $$[(a_n)] \preccurlyeq [(a_n)]$$

1. $$\preccurlyeq$$ is symmetric

Assume that $$[(a_n)] \preccurlyeq [(b_n)]$$ and $$[(b_n)] \preccurlyeq [(a_n)]$$.

It follows that $$\forall \epsilon >0, \exists N_1, \forall n>N_1: a_n - b_n < \epsilon$$ and $$\forall \epsilon >0, \exists N_2, \forall n>N_2: b_n - a_n < \epsilon$$.

Then $$\forall \epsilon >0, \exists N_1, \forall n>N_1: a_n - b_n < \epsilon$$ and $$\forall \epsilon >0, \exists N_2, \forall n>N_2: -\epsilon < a_n - b_n$$.

Take $$N=\max \{N_1,N_2\}$$.

Hence $$\forall \epsilon >0, \exists N, \forall n>N: |a_n - b_n| < \epsilon$$ and thus $$(a_n) \sim (b_n)$$. As a result, $$[(a_n)] = [(b_n)]$$.

1. $$\preccurlyeq$$ is transitive

Assume that $$[(a_n)] \preccurlyeq [(b_n)]$$ and $$[(b_n)] \preccurlyeq [(c_n)]$$.

It follows that $$\exists N_1, \forall n>N_1: a_n - b_n < \dfrac{\epsilon}{2}$$ and $$\exists N_2, \forall n>N_2: b_n - c_n < \dfrac{\epsilon}{2}$$.

Take $$N=\max \{N_1,N_2\}$$.

Hence $$\forall \epsilon >0, \exists N, \forall n>N: a_n - c_n = (a_n-b_n)+(b_n-c_n) <\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}= \epsilon$$.

As a result, $$[(a_n)] \preccurlyeq [(c_n)]$$.

1. $$\preccurlyeq$$ is linear

I presented a proof here.

• In argument 3 you should fix a rational epsilon $\varepsilon>0$ before you define $N$. Beyond what you written being cacographic, everything you wrote seems to have nothing wrong (that cannot be immediately fixed; e.g. my first remark). – Alberto Takase Feb 1 at 3:56