Metric Space Completeness is Preserved by Isometry/Proof 2

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

Let $\sequence {b_n}$ be a Cauchy sequence in $A_2$.

Thus:
 * $\exists N_1 \in \N: \map {d_2} {b_n, b_m} < \epsilon$

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:
 * $\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } {b_m} } = \map {d_2} {b_n, b_m} < \epsilon$

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

So $\sequence {\map {\phi^{-1} } {b_n} }$ is Cauchy in $A_1$.

Since $A_1$ is complete, $\sequence {\map {\phi^{-1} } {b_n} }$ 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 $\map {\phi^{-1} } b = a$.

Since $\sequence {\map {\phi^{-1} } {b_n} }$ converges to $\map {\phi^{-1} } b$, there exists some $N_2 \in \N$ such that:


 * $\map {d_1} {\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b} < \epsilon$

whenever $n \ge N_2$ and $\map {\phi^{-1} } {b_n}, \map {\phi^{-1} } b \in A_1$.

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

and so:
 * $\map {d_2} {b_n, b} < \epsilon$

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

Thus $\sequence {b_n}$ converges in $A_2$.

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