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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.

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  • $\begingroup$ 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). $\endgroup$ – Alberto Takase Feb 1 at 3:56

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