Isometric Image of Cauchy Sequence is Cauchy Sequence

Theorem
Let $\left({S_1,d_1}\right)$ and $\left({S_2,d_2}\right)$ be metric spaces.

Let $f: S_1 \to S_2$ be an isometry.

Let $\left\langle{x_n}\right\rangle$ be a Cauchy sequence in $S_1$.

Let $\left\langle{y_n}\right\rangle = \left\langle{f\left({x_n}\right)}\right\rangle$ be the image of $\left\langle{x_n}\right\rangle$ under $f$.

Then $\left\langle{y_n}\right\rangle$ is a Cauchy sequence.

Proof
Let $\epsilon \in \R_{>0}$.

By the definition of Cauchy sequence, there is an $N \in \R$ such that:


 * $\left({m > N}\right) \land \left({n > N}\right) \implies d_1 \left({x_m, x_n}\right) < \epsilon$

Since $f$ is an isometry, $d_2 \left({y_m, y_n}\right) = d_1 \left({x_m, x_n}\right)$ for all $m$ and $n$.

Thus:


 * $\left({m > N}\right) \land \left({n > N}\right) \implies d_2 \left({y_m, y_n}\right) < \epsilon$

Hence $\left\langle{y_n}\right\rangle$ is a Cauchy sequence.