## Completion of a metric space

Let $(X,d)$ be a metric space. We want to consider “Cauchy” sequences: sequences $\{ x_n \} \subseteq X$ whose terms get arbitrarily close together for sufficiently large indices $n$.

DefinitionA sequence $\{ x_n \} \subseteq X$ is Cauchy if for all $\epsilon > 0$ there exists a natural number $N$ such that for all $p, q \geq N$, we obtain $d(x_p,x_q) < \epsilon$.

In a sense, these sequences “want” to converge to something, so it seems reasonable on first glance to expect that they should do so. However, it turns out that there may be “gaps” in the metric space: there could be sequences which are somehow “converging to a limit point which isn’t there”. When there are no such gaps, we have a special name for the space.

Definition. A metric space $(X,d)$ is complete if every Cauchy sequence $\{ x_n \} \subseteq X$ is convergent.

A theoretical question now arises: is it possible, given an arbitrary metric space, to construct a “larger” space in which all of its gaps are filled in?

This leads directly to the notion of a completion of a metric space: we want a complete metric space $(Y,d_Y)$ which contains a copy of $(X,d_X)$ living inside of it, and is as small as possible. For example, if we consider the rational numbers $\mathbf{Q}$ as a metric space, then $\mathbf{R}$ and $\mathbf{C}$ are both complete metric spaces containing a copy of $\mathbf{Q}$, however $\mathbf{R}$ is the “smallest”, and so is of more interest to us.

In order to make this formal we need the notion of an isometry.

Definition. A function $F : (X,d_X) \to (Y,d_Y)$ is called an isometry if for all $x,y \in X$ we have $d_Y(F(x), F(y)) = d_X(x,y)$.

In this sense, an isometry is a very rigid map — we note at once that any such map must be injective, since it must preserve distinctness of points. We then have the following definition.

Definition. Let $(X,d_X)$ be a metric space. A metric space $(Y,d_Y)$ is called a completion of $(X,d_X)$ if it satisfies the following:

1. $(Y,d_Y)$ is complete.
2. There exists an isometry $\varphi : (X,d_X) \to (Y,d_Y)$.
3. $\overline{\varphi(X)} = Y$, that is, the closure of the “copy” of $X$ living inside $Y$ is the whole space.

In the same way the notions of convergence and Cauchyness carried over to general metric spaces, so too do those of continuity and uniform continuity. The completion of $(X,d_X)$ is characterized, up to a bijective isometry, by the following universal property: for every uniformly continuous map $f : (X,d_X) \to (Z,d_Z)$ there is a unique uniformly continuous map $g : (Y,d_Y) \to (Z,d_Z)$ that “extends” $f$, in the sense that $f(x) = g(\varphi(x))$ for all $x \in X$.

It seems, from my intuition, that the following characterization should also work (however I haven’t tried to prove this yet). It is based loosely on the following idea: if $(Z,d_Z)$ is any complete metric space with a copy of $(X,d_X)$ sitting inside it, then it should contain a copy of $(Y,d_Y)$ as well. This is stated formally below.

Conjecture. $(Y,d_Y)$ is a completion of $(X,d_X)$ (with the corresponding isometry $\varphi$) if and only if for any complete metric space $(Z,d_Z)$ and any isometry $\psi : (X,d_X) \to (Z,d_Z)$, there is a (unique?) isometry $\beta : (Y,d_Y) \to (Z,d_Z)$ such that $\beta \circ \varphi = \psi$.

just another guy trying to make the diagrams commute.
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### 6 Responses to Completion of a metric space

1. matto says:

“it turns out that there may be “gaps” in the metric space: there could be sequences which are somehow “converging to a limit point which isn’t there”.”

like a_n = [1; 1, …, 1] (with n ones) over Q; a_n → φ

• mlbaker says:

don’t understand your notation for $a_n$

• I believe that’s notation for a continued fraction.

2. Sorry to bump up an old post, but I was wondering was phi_X [bar] was. That is, what is the closure of the copy of X?

• mlbaker says:

$\phi(X)$ is a subset of $Y$. Are you asking how to define the closure of a subset $A$ of a metric space? It’s the union of $A$ with the set of $A$‘s limit points.

• So essentially, the smallest closed set which contains $\phi(X)$ ?