Isometric Image of Cauchy Sequence is Cauchy Sequence

Theorem
Let $(S_1,d_1)$ and $(S_2,d_2)$ be metric spaces.

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

Let $(x_n)$ be a Cauchy sequence in $S_1$.

Let $(y_n) = (f(x_n))$ be the image of $(x_n)$ under $f$.

Then $(y_n)$ is a Cauchy sequence.

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

By the definition of Cauchy sequence, there is an $N \in \R$ such that
 * $(m > N) \land (n > N) \implies d_1(x_m, x_n) < \epsilon$.

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

Thus
 * $(m > N) \land (n > N) \implies d_2(x_m, x_n) < \epsilon$.

Thus $(y_n)$ is a Cauchy sequence.