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

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

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

Let $$ \left(a_n\right),\left(b_n\right),\left(c_n\right)\in <!--\; \in \; -->{\mathcal{C}}_{\mathbb{Q}}$$, let's show that the relation is reflexive, first we have to show that $$ a_n-<!--\; -\; -->a_n\to <!--\; \to \; -->0$$ 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 $$ a_n-<!--\; -\; -->b_n\to <!--\; \to \; -->0$$, and we must show that $$ b_n-<!--\; -\; -->a_n\to <!--\; \to \; -->0$$, so let $$ \mathrm{ϵ<!--\; ϵ\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{+}$$, so we get some $$ N\in <!--\; \in \; -->{\mathbb{N}}_0$$ such that for all $$ n\ge <!--\; \ge \; -->{\mathbb{N}}_0$$ we have that $$ |<!--\; |\; -->a_n-<!--\; -\; -->b_n|<!--\; |\; --><<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->}$$ but since we can move things around inside of the absolute value bars we get $$ |<!--\; |\; -->b_n-<!--\; -\; -->a_n|<!--\; |\; --><<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->}$$ as needed.