Completion Theorem (Metric Space)/Lemma 4

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $\tilde M = \struct {\tilde A, \tilde d}$ be a completion of $\struct {A, d}$.

Then:
 * $\tilde M = \struct {\tilde A, \tilde d}$ is unique up to isometry.

That is, if $\struct {\hat A, \hat d}$ is another completion of $\struct {A, d}$, then there is a bijection $\tau: \tilde A \leftrightarrow \hat A$ such that:
 * $(1): \quad \tau$ restricts to the identity on $x$:
 * $\forall x \in A : \map \tau x = x$
 * $(2): \quad \tau$ preserves metrics:
 * $\forall x_1, x_2 \in A : \map {\hat d} {\map \tau {x_1}, \map \tau {x_2} } = \map {\tilde d} {x_1, x_2}$

Proof
Let $M_1 = \struct {\tilde{A_1}, \tilde{d_1}, \phi_1}$ and $M_2 = \struct {\tilde{A_2}, \tilde{d_2}, \phi_2}$ be two completions of $\struct {A, d}$.

Here, $\phi_1: A \to A_1$ and $\phi_2: A \to A_2$ are isometries

By Composite of Isometries is Isometry, $\psi = \phi_1^{-1} \circ \phi_2$ gives an isometry from $\phi_1 \sqbrk A$ to $\phi_2 \sqbrk A$.

The sets $\phi_1 \sqbrk A$ and $\phi_2 \sqbrk A$ are dense in $A_1$ and $A_2$ respectively.

Thus we can extend $\psi$ continuously to a map $\psi: A_1 \to A_2$.

That is, for $x \in A_1$, we can find a Cauchy sequence $\sequence {w_n}$ in $A_1$ with limit $x$.

Then we define:
 * $\displaystyle \map \psi x := \lim_{n \mathop \to \infty} \map \psi {w_n}$

which converges as $A_2$ is complete.

By Metric Space is Hausdorff, $A$ is a Hausdorff space.

By $T_2$ (Hausdorff) Space is Preserved under Homeomorphism, $A_1$ and $A_2$ are also Hausdorff.

Therefore, by Convergent Sequence in Hausdorff Space has Unique Limit, $\psi$ is well-defined.

It is to be shown that $\psi$ is surjective:

For $y \in \tilde {A_2}$, let $\sequence {w_n'}$ be a Cauchy sequence in $\phi_2 \sqbrk A$ with limit $y$ in $\tilde {A_2}$.

Let $z_n$ be the preimage of the $w_n'$ under $\psi$.

Then, as $A_2$ is Hausdorff:
 * $\displaystyle \lim_{n \mathop \to \infty} \map \psi {z_n} = y$

as required.

injectivity of $\psi$ holds because $A_1$ is Hausdorff, as follows:

Suppose that:
 * $\displaystyle \lim_{n \mathop \to \infty} \map \psi {w_n} = \lim_{n \mathop \to \infty} \map \psi {w_n'}$
 * $\displaystyle \lim_{n \mathop \to \infty} w_n = w$

and:
 * $\displaystyle \lim_{n \mathop \to \infty} w_n' = w'$.

For $\epsilon > 0$, pick $M \in \N$ such that $\map \psi {w_n}$, $\map \psi {w_n'}$ lie in the open $\epsilon$-ball $\map {B_{\epsilon / 2} } {\map \psi w}$ for all $n \ge M$.

Then we have:


 * $\map {\tilde {d_1} } {w_n, w_n'} = \map {\tilde {d_2} } {\map \psi {w_n}, \map \psi {w_n'} } \le \epsilon$

As $A_1$ is Hausdorff, we conclude $w = w'$, and the task is complete.

Finally, from Distance Function of Metric Space is Continuous, it follows that $\psi$ is an isometry on all of $A_1$, and the proof is complete.