We want to show that the set \(I:=\{(a_n)_{n\in\mathbb N}~|~a_n\in\mathbb Q,\lim a_n=0\}\) of all rational sequences convergent to \(0\) is a subgroup of the commutative group of rational Cauchy sequences \((M, + )\), formally \[(I, + )\subseteq (M, + ).\]
We will do so by applying the first subgroup criterion:
\((1)\) Note that \(I\) is not empty, since it contains the rational sequence consinsting of rational zeros \((0)_{n\in\mathbb N}\), because \(\lim 0=0\).
\((2)\)
Moreover, we have already proven that \(I\) is a subset of \(M\), formally \(I\subseteq M\). Therefore, for any rational sequences \((a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\) being elements of \(I\) we can conclude by definition of adding rational Cauchy sequences, that
\[(a_n)_{n\in\mathbb N} + [-(b_n)_{n\in\mathbb N}]=(a_n-b_n)_{n\in\mathbb N}\in M,\]
which means that the rational sequence \((a_n-b_n)_{n\in\mathbb N}\) is en element of \(M\), i.e. it is a rational Cauchy sequence.
\((3)\) It remains to be shown that \((a_n-b_n)_{n\in\mathbb N}\) is an element of \(I\), or \(\lim a_n-b_n=0\). By hypothesis, \((a_n)_{n\in\mathbb N},(b_n)_{n\in\mathbb N}\in I\), which is equivalent to \(\lim a_n=0\) and \(\lim b_n=0\). Following the definition of convergence in the metric space \((\mathbb Q,|~|)\), for an arbitrarily small \(\epsilon/2\), \(\epsilon\in\mathbb Q\) there exist two indices \(N_a(\epsilon/2), N_b(\epsilon/2), \in\mathbb N\) such that for all \(n > \max (N_a(\epsilon/2),N_b(\epsilon/2))\), we have \[|a_n| < \frac \epsilon2\quad\text{and}\quad|b_n| < \frac \epsilon2.\]
Using the triangle inequality of the metric \(|\cdot|\)), we can estimate
\[|a_n-b_n|\le |a_n|+|b_n| < \frac\epsilon2 + \frac\epsilon2 = \epsilon.\]
This shows, that \(\lim a_n-b_n=0\), or that \((a_n-b_n)_{n\in\mathbb N}\in I\), or that \((I, +)\) is a subgroup of \((M, + )\).
