Metric Space Completeness is Preserved by Isometry/Proof 2

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

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

Thus:
 * $\exists N_1 \in \N: d_2 \left({b_n, b_m}\right) < \epsilon$

whenever $n, m \ge N_1$ and $b_n, b_m \in A_2$.

We have that $M_1$ is isometric to $M_2$.

Isometry is Equivalence Relation and so in particular symmetric.

Hence $M_2$ is isometric to $M_1$, via $\phi^{-1}$.

Thus:
 * $d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b_m}\right)}\right) = d_2 \left({b_n, b_m}\right) < \epsilon$

whenever $n, m \ge N_1$ and $\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b_m}\right) \in A$.

So $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ is Cauchy in $A_1$.

Since $A_1$ is complete, $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ converges in $A_1$ to, say, $a$.

By definition of isometry, $\phi^{-1}$ is a bijection, and in particular surjective

Thus there exists some $b \in A_2$ such that $\phi^{-1} \left({b}\right) = a$.

Since $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ converges to $\phi^{-1} \left({b}\right)$, there exists some $N_2 \in \N$ such that:


 * $d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right)}\right) < \epsilon$

whenever $n \ge N_2$ and $\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right) \in A_1$.

Since $M_2$ is isometric to $M_1$, we have:
 * $d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right)}\right) = d_2 \left({b_n, b}\right)$

and so:
 * $d_2 \left({b_n, b}\right) < \epsilon$

whenever $n \ge N_2$ and $b_n, b \in A_2$.

Thus $\left\langle{b_n}\right\rangle$ converges in $A_2$.

Since $\left\langle{b_n}\right\rangle$ was an arbitrary Cauchy sequence, we have that $M_2$ is complete, as required.