🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Let $ϵ\in {\mathbb{R}}^{+}$ since $\left(a_n\right)$ converges to $q$ then we know that there exists some $M\in {\mathbb{N}}_0$ such that for any $n\ge N$ we have that $|a_n-L|<\frac{ϵ}{2}$, take $N=M$ and let $n,m\ge M$ then we have $${\displaystyle |a_n-a_m|=|(a_n-L)-(a_m-L)|\le |a_n-L|+|a_m-q|<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ}$$ where we've used the triangle inequality

Since $\left(a_n\right)$ is cauchy then with $ϵ=1$ we get an $N\in {\mathbb{N}}_0$ such that for any $m,n\ge N$ we have that $|a_n-a_m|<1$. Specificially since $N+1\ge N$ then we would know that $|a_{N+1}-a_m|\le 1$ which is the same as $${\displaystyle a_{N+1}-1<a_n<a_{N+1}+1}$$ so that $\left|a_n\right|\le max\left(\right|a_{N+1}-1|,|a_{N+1}+1\left|\right)$

Therefore set $${\displaystyle M=max\left(\right|a_0|,|a_1|,\dots ,|a_N|,|a_{N+1}-1|,|a_{N+1}+1\left|\right)}$$ With this, let $k\in {\mathbb{N}}_0$ then we know that if $k\le N$ then $\left|a_k\right|\le M$ as they are directly included in our maximum, on the other hand if $k>N$ then $\left|a_k\right|<max\left(\right|a_{N+1}-1|,|a_{N+1}+1\left|\right)\le M$ and therefore $\left(a_n\right)$ is bounded.

Let $\left(a_n\right),\left(b_n\right),\left(c_n\right)\in {\mathcal{C}}_{\mathbb{Q}}$, let's show that the relation is reflexive, first we have to show that $a_n-a_n\to 0$ the sequence $a_n-a_n$ is zero for every index so it trivially tends to 0.

To show that the relation is symmetric we assume that $a_n-b_n\to 0$, and we must show that $b_n-a_n\to 0$, so let $ϵ\in {\mathbb{R}}^{+}$, so we get some $N\in {\mathbb{N}}_0$ such that for all $n\ge {\mathbb{N}}_0$ we have that $|a_n-b_n|<ϵ$ but since we can move things around inside of the absolute value bars we get $|b_n-a_n|<ϵ$ as needed.