# completion of metric spaces

It is well known that every metric space has a completion. That is, up to isometry there is exactly one smallest complete space containing a given metric space as a subset. The way that that completion is normally found is by forming a quotient of the space of all Cauchy sequences of the original space. Even though this construction is quite straightforward there are some problems

- Spelling out all details is actually quite tedious (as I’ve noticed when I presented it to a class last year).
- It’s not the most elegant solution, in particular the quotient seems to be hard for students to work with. Having said that, it is actually fairly natural to consider all Cauchy sequences: every CS should converge, so we need some point it converges to, and we might as well use the CS as its label. Of course that is too many names, so we need to quotient.

There is a nicer construction that is less known. This construction seems to come from Fred Richman. We first need to define a central notion:

Consider a metric space $(X,d)$. A function $f:X^2 \to \mathbb{R}$ is called a

locationif \begin{equation*} f(x)-f(y) \leqslant d(x,y) \leqslant f(x)+f(y) \ , \end{equation*} and $\quad \inf f = 0$.

It is obvious, that for any point $x \in X$ the function $d(x,\cdot)$ is a location. In fact the set of all locations $L$ will be our completion. To define a metric on the set $L$, we define, for $f,g \in L$, \begin{equation} D(f,g) = \sup_{x \in X} d(f(x),g(x)) \end{equation}

Below is an outline of the rest of the construction. Notice that the paragraphs can be **expanded for more details**.

The biggest obstacle is to show that this is well defined. Interestingly we can even do this constructively. One can show that for any $f,g \in L$ and any $\varepsilon > 0$ there is $x_0 \in X$ such that for all $x \in X$ we have [ |f(x)-g(x) | \leqslant g(x_0) + \varepsilon \ . ] In words: for any given accuracy we can determine $d(f,g)$ by only considering the difference of the functions at one point. It might seem weird, that only $g$ is involved in this estimate, but the $x_0$ is chosen as an approximate root for $f$, so $f$ does play a role.

Choose $x_0$ such that $f(x_0) \lt \varepsilon$ (remember that $\inf f = 0$). Then \begin{align} f(x)-g(x) & = f(x) - f(x_0) - (g(x_0) + g(x)) + f(x_0) + g(x_0) \\\\ & \leqslant d(x,x_0) - d(x,x_0) + f(x_0) + g(x_0) \\\\ & = f(x_0) + g(x_0) \end{align} and \begin{align} g(x) -f(x) & = g(x) - g(x_0) - (f(x_0) + f(x)) + g(x_0) + f(x_0) \\ & \leqslant d(x,x_0) - d(x,x_0) + f(x_0) + g(x_0) \\\\ & = f(x_0) + g(x_0) \ . \end{align} Together \begin{align} |f(x) - g(x)| \leqslant g(x_0)+\varepsilon \ . \end{align}

Showing that $d$ is actually a metric is straightforward

Reflexiveness and symmetry are trivial. The triangle equality can be established in the same way (normally a good exercise for students) that the supremum metric on a bounded metric space is a metric.

As already mentioned above we can embed $X$ in $L$ by identifying $x$ with $d(x,\cdot)$, and it is again straightforward to show that this embedding preserves the metric.

Let $x,y \in X$. Then \begin{equation} D(d(x,\cdot),d(y,\cdot)) = \sup_{z \in X} |d(x,z) - d(y,z)| \leqslant |d(x,y)| \end{equation} But also \begin{equation} D(d(x,\cdot),d(y,\cdot)) = \sup_{z \in X} |d(x,z) - d(y,z)| \geqslant |d(x,y) - d(y,y)| = |d(x,y)| \ . \end{equation} So together $D(d(x,\cdot),d(y,\cdot)) = d(x,y)$.

Finally we need to show that $L$ is complete.

If $f_n$ is a Cauchy sequence in $L$ then it converges point-wise to a function $f:X \to \mathbb{R}$. This function is obviously the limit of $f_n$ even under the metric $D$. It is also easily seen to be a location again.