Metric Completion

A metric completion of a metric space is a complete metric space containing a distance-preserving copy of the original space as a dense subset. It supplies limits for all Cauchy sequences without changing distances between existing points or adding points unrelated to such limits. Every metric space has a completion, and this completion is unique up to an isometry that respects the embedding of the original space. (cis.upenn.edu)

Definition and motivation

For a metric space $(X,d)$, a completion consists of a complete metric space $(\widehat X,\widehat d)$ and an isometric embedding

such that

The bar denotes closure. Density means that every point of $\widehat X$ is a limit of points from $i(X)$. (cis.upenn.edu)

A Cauchy sequence has terms that become arbitrarily close to one another, but its limit need not belong to the space. For example, rational approximations to $\sqrt2$ form a Cauchy sequence in the rational numbers with their usual distance, although they have no rational limit. The real numbers provide the missing limits. Completeness requires convergence of every Cauchy sequence, not of every sequence. (jirka.org)

Construction from Cauchy sequences

Let $\mathcal C(X)$ be the set of Cauchy sequences in $X$. Define an equivalence relation by

The completion is the set of equivalence classes

with distance

This limit exists because

so the real-valued sequence of distances is Cauchy. The triangle inequality also shows that changing representatives does not change the limit. Taking equivalence classes ensures that distinct completed points have positive distance. (math.gsu.edu)

The embedding sends $x$ to the class of the constant sequence $(x,x,\ldots)$. For every Cauchy sequence $(x_n)$, the points $i(x_n)$ converge to $[(x_n)]$, establishing density. Completeness follows by approximating the terms of a Cauchy sequence in $\widehat X$ by points of $i(X)$, with errors tending to zero, and using the resulting Cauchy sequence in $X$. Thus the construction works without an already available ambient complete space. (math.gsu.edu)

Uniqueness and extension of maps

If $(\widehat X_1,i_1)$ and $(\widehat X_2,i_2)$ are completions of $X$, there is exactly one surjective isometry

Its values are forced by

Uniqueness therefore concerns the completion together with its embedding, rather than a particular set-theoretic representation. (math.uwaterloo.ca)

The associated universal property is an extension theorem: every uniformly continuous map $f:X\to Y$, where $Y$ is complete, extends uniquely to a uniformly continuous map

Uniform continuity ensures that images of Cauchy sequences are Cauchy and that equivalent sequences yield the same limit. If $f$ satisfies Lipschitz continuity with constant $L$, passing its distance inequality to limits shows that $\widehat f$ has the same constant. (math.uwaterloo.ca)

Examples and analytical applications

If $X$ is isometrically embedded in a complete space $Y$, its closure in $Y$ is a completion. Consequently, $\mathbb Q$ completes to $\mathbb R$, and the interval $(0,1)$, with usual distance, completes to $[0,1]$. An already complete space acquires no new points: its image in any completion is both dense and closed. (jirka.org)

In functional analysis, completion preserves compatible linear structure. Every normed vector space completes to a Banach space, with vector operations defined on sequence representatives and extended by limits. Likewise, an inner-product space completes to a Hilbert space. (ocw.mit.edu)

A concrete example is the space of polynomials on a closed bounded interval, equipped with the uniform norm. Its completion is the space of continuous functions on that interval: polynomial approximation gives density, while uniform limits of continuous functions remain continuous. Completion thus turns a simpler class of approximating objects into a space containing all their admissible limits. (math.gsu.edu)

Dependence on the metric and limitations

Completion depends on the metric, not merely on the induced topology. For example, on $X=(0,1)$, consider

Both induce the usual topology. However, $d$ completes to $[0,1]$, whereas $\rho$ is already complete: the displayed logarithmic map is an isometry onto $\mathbb R$. This directly illustrates that topologically equivalent metrics can have different Cauchy sequences and completions. (jirka.org)

Ordinary continuity is insufficient for the extension theorem. For instance, $f(x)=1/x$ is continuous on $(0,1)$, but cannot extend continuously to its usual completion $[0,1]$, because it has no finite limit at zero. This example shows why control over Cauchy sequences matters. (math.uwaterloo.ca)

Completion is also distinct from making a space compact. The completion of $\mathbb Q$ is noncompact $\mathbb R$. More generally, a completion is a compact space exactly when the original space is totally bounded: completeness alone does not provide the finite-covering property needed for compactness. (jirka.org)

References

1. 9. Completion of a Metric Spacecis.upenn.edu
2. Functional Analysis Notesmath.gsu.edu
3. PMath 351 Notesmath.uwaterloo.ca
4. RA Completeness and compactnessjirka.org
5. Functional Analysis Lecture Notes, Spring 2020ocw.mit.edu