Proof
(related to Lemma: Convergent Sequences are Cauchy Sequences (Metric Spaces))
- By hypothesis, \((a_n)_{n\in\mathbb N}\) is a convergent sequence of points in the mectric space \((X,d)\) with a limit \(\lim_{n\rightarrow\infty} a_n=a\).
Thus, for each \(\epsilon > 0\) there exists an index \(N\in\mathbb N\) with \[d(a_n,a) < \epsilon\quad\quad\text{ for all }n\ge N.\]
- Therefore, and because of the property of the metric (triangle inequality) , it follows for all \(i,j\ge N\) \[ d(a_i,a_j) \le d(a_i,a) + d(a,a_j) < \epsilon + \epsilon =2\epsilon.\]
- Since \(\epsilon > 0\) can be chosen arbitrarily close to \(0\), it follows that the sequence \((a_n)\) is also a Cauchy sequence.
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References
Bibliography
- Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984
