Metric Space Completeness is Preserved by Isometry/Proof 1

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $\phi: M_1 \to M_2$ be an isometry.

If $M_1$ is complete then so is $M_2$.

Proof
Let $\tau_1$ be the topology on $A_1$ induced by $d_1$.

Let $\tau_2$ be the topology on $A_2$ induced by $d_2$.

Let $\left\langle{x_n}\right\rangle$ be a Cauchy sequence in $A_2$.

From Inverse of Isometry is Isometry, $\phi^{-1}$ is an isometry.

Since Isometric Image of Cauchy Sequence is Cauchy Sequence, $\left\langle{\phi^{-1} \left({x_n}\right)}\right\rangle$ is a Cauchy sequence.

Since $M_1$ is a |complete metric space, $\left\langle{\phi^{-1} \left({x_n}\right)}\right\rangle$ converges.

Since Isometry Preserves Sequence Convergence, $\left\langle{\phi \left({\phi^{-1} \left({x_n}\right)}\right)}\right\rangle$ converges.

But:
 * $\left\langle{\phi \left({\phi^{-1} \left({x_n}\right)}\right)}\right\rangle = \left\langle{x_n}\right\rangle$

so $\left\langle{x_n}\right\rangle$ converges.

Thus each Cauchy sequence in $M_2$ converges.

It follows by definition that so $M_2$ is a complete metric space.