User:Omlett71/Sandbox/Equivalence Relation on Rational Cauchy Sequences

Motivation
Bla bla bla

Group structure of Rational Cauchy Sequences
If $\langle x_n \rangle$ and $\langle y_n \rangle$, the sequence defined by adding them pointwise, denoted $\langle x_n \rangle + \langle y_n \rangle$ is also rational and Cauchy.

It is obviously rational because the sum of any two rational numbers is also rational. If $\epsilon \in \Q_{>0}$, then $\epsilon/2 \in \Q_{>0}$ and because $\langle x_n \rangle$, $\langle y_n \rangle$ are Cauchy, there must exist $N_x$ and $N_y$ satisfying $n,m> N_x \implies |x_n - x_m|<\epsilon/2 $ and $n,m> N_y \implies |y_n - y_m|< \epsilon/2$. If $N:=\max\{N_y,N_x\}$, then for $n,m>N$, we have $|(x_n+y_n)-(x_m+y_m)| \le |x_n - x_m| + |y_n - y_m| < \epsilon/2 + \epsilon/2 = \epsilon$ by the triangle inequality, and thus this sequence is Cauchy.

This operation is associative and commutative by the associativity and commutativity of $\struct{\Q,+}$.

Similarly, if $\langle x_n \rangle$ is a rational Cauchy sequence, $-\langle x_n \rangle$ defined by multiplying each element by -1 is also Cauchy. We simply note that $|(-x_n) - (-x_m)| = |-(x_n - x_m)| = |x_n - x_m|$, and so if $\epsilon \in \Q_{>0}$, then the $N$ satisfying $n,m > N \implies |x_n - x_m| < \epsilon$ also satisfies $n,m > N \implies |(-x_n)-(-x_m)|<\epsilon$.

Finally, the zero sequence is clearly rational Cauchy, and clearly an identity w.r.t. pointwise addition. We have that $\langle x_n \rangle + -\langle x_n \rangle =0$.

Lemma
We define the relation $\sim$ by $\langle x_n \rangle \sim \langle y_n \rangle \iff $\forany \epsilon \in \Q_{>0}:\exists