ThmDex – An index of mathematical definitions, results, and conjectures.
Result R244 on D65: Cauchy sequence
Convergent sequence is Cauchy

Let $M = (X, \mathcal{T}_d, d)$ be a D1107: Metric space such that
 (i) $x : \mathbb{N} \to X$ is a D336: Convergent sequence in $M$
Then $x$ is a D65: Cauchy sequence in $M$.

Let $M = (X, \mathcal{T}_d, d)$ be a D1107: Metric space such that
 (i) $\textsf{Convergent}(M)$ is the D2505: Set of convergent sequences in $M$ (ii) $\textsf{Cauchy}(M)$ is the D724: Set of Cauchy sequences in $M$
Then $$\textsf{Convergent}(M) \subseteq \textsf{Cauchy}(M)$$
Proofs
Proof 0
Let $M = (X, \mathcal{T}_d, d)$ be a D1107: Metric space such that
 (i) $x : \mathbb{N} \to X$ is a D336: Convergent sequence in $M$
Fix $\varepsilon > 0$. Since $x$ is convergent in $M$, then result R1089: Characterisation of convergent sequences in metric space shows that there is a point $a \in X$ and a threshold $N \in \mathbb{N}$ such that $d(x_n, a) < \varepsilon / 2$ for every $n \geq N$. Thus, if $n_0, n_1 \in \mathbb{N}$ such that $n_0, n_1 \geq N$, then $$d(x_{n_0}, x_{n_1}) \leq d(x_{n_0}, a) + d(a, x_{n_1}) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$ Since $\varepsilon > 0$ was arbitrary, the proof is complete. $\square$