# Proof

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 $$(M, + )$$.

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### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013