Completion Theorem (Metric Space)/Lemma 4

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $\tilde M = \left({\tilde A, \tilde d}\right)$ be a completion of $\left({A, d}\right)$.

Then:
 * $\tilde M = \left({\tilde A, \tilde d}\right)$ is unique up to isometry.

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

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

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 $\phi_1 \left({A}\right)$ to $\phi_2 \left({A}\right)$.

The sets $\phi_1 \left({A}\right)$ and $\phi_2 \left({A}\right)$ 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 $\left\langle{w_n}\right\rangle$ in $A_1$ with limit $x$.

Then we define:
 * $\displaystyle \psi \left({x}\right) := \lim_{n \mathop \to \infty} \psi \left({w_n}\right)$

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 $\left\langle{w_n'}\right\rangle$ be a Cauchy sequence in $\phi_2 \left({A}\right)$ 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:
 * $\displaystyle \lim_{n \mathop \to \infty} \psi \left({z_n}\right) = y$

as required.

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

Suppose that:
 * $\displaystyle \lim_{n \mathop \to \infty} \psi \left({w_n}\right) = \lim_{n \mathop \to \infty} \psi \left({w_n'}\right)$
 * $\displaystyle \lim_{n \mathop \to \infty} w_n = w$

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

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

Then we have:


 * $\tilde {d_1} \left({w_n, w_n'}\right) = \tilde {d_2} \left({\psi \left({w_n}\right), \psi \left({w_n'}\right)}\right) \le \epsilon$

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

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