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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces