# Metric Space Completeness is Preserved by Isometry

## Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

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

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

## Proof 1

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 $\sequence {x_n}$ be a Cauchy sequence in $A_2$.

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

Since Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\phi^{-1} } {x_n} }$ is a Cauchy sequence.

Since $M_1$ is a complete metric space, $\sequence {\map {\phi^{-1} } {x_n} }$ converges.

Since Isometry Preserves Sequence Convergence, $\sequence {\map \phi {\map {\phi^{-1} } {x_n} } }$ converges.

But:

$\sequence {\map \phi {\map {\phi^{-1} } {x_n} } } = \sequence {x_n}$

so $\sequence {x_n}$ converges.

Thus each Cauchy sequence in $M_2$ converges.

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

$\blacksquare$

## Proof 2

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.

$\blacksquare$