A completion of a metric space $(M,d)$ is a complete metric space $(M^*,d^*)$ such that $(M,d)$ is a dense subspace of $(M^*,d^*)$. I understand this, but how do I explicitely find a completion of a given space?

I need to find the completion of the non complete metric space of real valued compact supported continuous functions with sup norm but I have no idea on how to. I already proved that the space is not complete using the hints in these two answers:

Is the set of all real valued continuous functions on $\mathbb R$ with compact support complete?

Completeness of continuous real valued functions with compact support



The classic thing here is the Cauchy construction. You have your metric space $M$, and you take the set $X$ of Cauchy sequences of $M$. Now put an equivalence relation by saying two Cauchy sequences $x_i$ and $y_i$ are equivalent if $d(x_i,y_i)\to 0$. Call the set of equivalence classes $M'$, and put a metric on $M'$ by saying that $d([x_i],[y_i]) = \lim_{i\to \infty} d(x_i,y_i)$, and then you have a map $M\to M'$ by $x\mapsto [(x,x,\ldots)]$.

You'll have to check that this construction works. First that the relation is an equivalence relation, then that the metric is well-defined (existence of the limit + answer is independent of choice of representative), and then that this gives a complete metric space.

The intuition here is that "holes" in $M$ are really just Cauchy sequences with no limit, so if you're trying to define a new point that goes in that hole, define it to be that Cauchy sequence.

Another method which I like is to embed $M$ isometrically into $C_b(M)$ (the space of bounded continuous functions on $M$, with sup norm) by fixing $x_0 \in M$ and defining $g_x(y) = d(x,y) - d(x_0,y)$, then the map $x\mapsto g_x$ is our embedding. $C_b(M)$ is a Banach space, so then we just take the closure of the image.


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