(related to Lemma: Convergent Rational Sequences With Limit \(0\) Are Rational Cauchy Sequences)
Let \((M , + , \cdot)\) be the unit ring of all rational Cauchy sequences. We want to show that the set \(I:=\{(a_n)_{n\in\mathbb N}~|~a_n\in\mathbb Q,\lim a_n=0\}\), i.e. the set of all rational sequences convergent to \(0\) is a subset of \(M\), formally \(I\subseteq R\). We will demonstrate that any rational sequence \((a_n)_{n\in\mathbb N}\) with \(\lim a_n=0\) is also a rational Cauchy sequence, formally \[(a_n)_{n\in\mathbb N}\in I\Longrightarrow (a_n)_{n\in\mathbb N}\in M.\]
Following the definition of convergence in the metric space \((\mathbb Q,|~|)\), for any arbitrarily small \(\epsilon/2\), \(\epsilon\in\mathbb Q\), there exists an \(N(\epsilon/2)\in\mathbb N\) such that for all \(n , m > N(\epsilon/2)\), we have \[|a_n| < \frac \epsilon2\quad\text{and}\quad|a_m| < \frac \epsilon2.\]
Thus, we can estimate (using the triangle inequality, see third property of the metric \(|\cdot|\)),
\[|a_n-a_m|\le |a_n|+|a_m| < \frac\epsilon2 + \frac\epsilon2 = \epsilon.\]
This demonstrates, that \((a_n)_{n\in\mathbb N}\) with \(\lim a_n=0\) is also a rational Cauchy sequence, or that \(I\subseteq M\), as required.
