Convergent Sequence of Rational Numbers

October 16, 2024

This is as proof that a convergent sequence of rational numbers is a Cauchy sequence.

Be (an)⊂Q(a_n) \subset \mathbb{Q} that converges to kk, then for each ϵ2>0\frac{\epsilon}{2} > 0, there exists n0∈Nn_0 \in \mathbb{N} such that:

∀n∈N,n≥n0  ⟹  ∣an−k∣<ϵ2 \forall n \in \mathbb{N}, n \geq n_0 \implies |a_n - k| < \frac{\epsilon}{2}

Now, for m,n∈Nm, n \in \mathbb{N}, m≥n0m \geq n_0 and n≥n0n \geq n_0, we have:

∣am−k∣<ϵ2and∣an−k∣<ϵ2 |a_m - k| < \frac{\epsilon}{2} \quad \text{and} \quad |a_n - k| < \frac{\epsilon}{2}

By the triangle inequality, we have:

∣am−an∣=∣(am−k)+(k−an)∣≤∣am−k∣+∣k−an∣=∣am−k∣+∣an−k∣<ϵ2+ϵ2=ϵ \begin{aligned} |a_m - a_n| & = |(a_m - k) + (k - a_n)| \\ & \leq |a_m - k| + |k - a_n| \\ & = |a_m - k| + |a_n - k| \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{aligned}

Thus, (an)(a_n) is a Cauchy sequence.