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 $M$.

Result R244
on D65: Cauchy sequence

Convergent sequence is Cauchy

Formulation 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 $M$ |

Then $x$ is a D65: Cauchy sequence in $M$.

Formulation 1

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
\begin{equation}
\textsf{Convergent}(M)
\subseteq \textsf{Cauchy}(M)
\end{equation}

Proofs

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
\begin{equation}
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
\end{equation}
Since $\varepsilon > 0$ was arbitrary, the proof is complete. $\square$