Completion Theorem (Metric Space)

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

Then there exists a completion $\tilde M = \left({\tilde A, \tilde d}\right)$ of $\left({A, d}\right)$.

Moreover, this completion 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
We construct the completion of a metric space as equivalence classes of the set of Cauchy sequences in the space under a suitable equivalence relation.

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

Let $\mathcal C \left[{A}\right]$ denote the set of all Cauchy sequences in $A$.

Define a relation $\sim$ on $\mathcal C \left[{A}\right]$ by:


 * $\displaystyle \left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle \iff \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right) = 0$

By Equivalence Relation on Cauchy Sequences, $\sim$ is an equivalence relation on $\mathcal C \left[{A}\right]$.

Denote the equivalence class of $\left\langle{x_n}\right\rangle \in \mathcal C \left[{A}\right]$ by $\left[{x_n}\right]$.

Denote the set of equivalence classes under $\sim$ by $\tilde A$.

By Relation Partitions Set iff Equivalence this is a partition of $\mathcal C \left[{A}\right]$.

That is, each $\left\langle{x_n}\right\rangle \in \mathcal C \left[{A}\right]$ lies in one and only one equivalence class under $\sim$.

Define $\tilde d: \tilde A \to \R_{\ge 0}$ by:


 * $\displaystyle \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right)$

Lemma 1: $\tilde d$ is Well-Defined
We claim that $\left({\tilde A, \tilde d}\right)$ is a completion of $\left({A, d}\right)$.

Therefore we must show that:
 * $\tilde d$ is a metric on $\tilde A$
 * There exists an everywhere dense inclusion $\left({A, d}\right) \to \left({\tilde A, \tilde d}\right)$ preserving $d$.

In addition the theorem claims that $\left({\tilde A, \tilde d}\right)$ is unique up to isometry.