\begin{definition}[Definition:Completion (Metric Space)]
Let $M_1 = \struct {A, d}$ and $M_2 = \struct {\tilde A, \tilde d}$ be metric spaces.
Then $M_2$ is a '''completion''' of $M_1$, or $M_2$ '''completes''' $M_1$, {{iff}}:
:$(1): \quad M_2$ is a complete metric space
:$(2): \quad A \subseteq \tilde A$
:$(3): \quad A$ is dense in $M_2$
:$(4): \quad \forall x, y \in A : \map {\tilde d} {x, y} = \map d {x, y}$. In terms of restriction of functions, this says that $\map {\tilde d {\restriction_A} } = d$.
It is immediate from this definition that a '''completion''' of a metric space $M_1$ consists of:
:A complete metric space $M_2$
:An isometry $\phi : A \to \tilde A$
such that $\map \phi A = \set {\map \phi x: x \in A}$ is dense in $M_2$.
An isometry is often required to be bijective, so here one should consider $\phi$ as a mapping from $A$ to the image of $\phi$.
Therefore to insist that $\phi$ be an isometry, ''in this context'', is to say that $\phi$ must be an injection that preserves the metric of $M_1$.
\end{definition}