Completion Theorem (Metric Space)

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

Then there exists a completion $\tilde M = \struct {\tilde A, \tilde d}$ of $\struct {A, d}$.

Moreover, this completion 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
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 $\struct {A, d}$ be a metric space.

Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$.

Define a relation $\sim$ on $\CC \sqbrk A$ by:


 * $\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$

By Equivalence Relation on Cauchy Sequences, $\sim$ is an equivalence relation on $\CC \sqbrk A$.

Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\sqbrk {x_n}$.

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

By Relation Partitions Set iff Equivalence this is a partition of $\CC \sqbrk A$.

That is, each $\sequence {x_n} \in \CC \sqbrk A$ lies in one and only one equivalence class under $\sim$.

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


 * $\ds \map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$

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

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

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