Definition:Completion (Metric Space)

From ProofWiki
Jump to navigation Jump to search


Let $M_1 = \left({A, d}\right)$ and $M_2 = \left({\tilde A, \tilde d}\right)$ be metric spaces.

Then $M_2$ is a completion of $M_1$, or $M_2$ completes $M_1$, if and only if:

$(1): \quad M_2$ is a complete metric space
$(2): \quad A \subseteq \tilde A$
$(3): \quad A$ is dense in $M_2$
$(4): \quad \forall x, y \in A : \tilde d \left({x, y}\right) = d \left({x, y}\right)$. In terms of restriction of functions, this says that $\displaystyle \tilde d {\restriction_A} = d$.

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 $\phi \left({A}\right) = \left\{ {\phi \left({x}\right) : x \in A} \right\}$ 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 $M_1$.