I'm a little embarrassed to ask this one, but it could help for a class I'm teaching, so here goes:

Let $X$ be a metric space. We all know that $X$ admits a completion, which is a complete metric space $\hat{X}$ together with an isometric embedding $\iota: X \hookrightarrow \hat{X}$ with dense image. Moreover, one learns that this completion is essentially unique.

From a modern perspective, one would like to realize the completion as satisfying some universal mapping property: this makes precise the "essentially unique" above and gives some functorial properties. But it seems to me that the completion satisfies two different such properties:

1) It is universal with respect to isometric embeddings into complete metric spaces.

2) It is universal with respect to uniformly continuous maps into complete metric spaces.

Of course 1) is the more obvious one. I gather from some internet research that 2) is supposed to be the "right" choice, and its usefulness is related to the fact that uniformly continuous maps have the extension property (again, I don't quite remember this from my undegraduate days; is it in Rudin's Principles, for instance?). However, it seems strange to me that by taking 2), we also get for free that the mapping $\iota$ is an isometric embedding (in particular, from 2) it doesn't even seem completely obvious that it is injective). Certainly one can see this by constructing the completion, but is there a more direct way?

I suspect that this is an instance when the more categorical thinkers have one up on me, and I stand ready to be enlightened.

  • $\begingroup$ @unknown: Thanks for the reference. I do know the result now. (Actually I learned it when teaching an undergraduate analysis class a few years back. It was covered in Russell Gordon's text.) I was making more of an expository point: that I don't remember this key fact being discussed in the basic analysis courses when I first learned about completions of metric spaces. $\endgroup$ – Pete L. Clark Jan 14 '10 at 7:14
  • $\begingroup$ @unknown: Apologies are unnecessary. Your comment was helpful. $\endgroup$ – Pete L. Clark Jan 15 '10 at 0:05
  • $\begingroup$ [deleted previous comments] $\endgroup$ – user2734 May 17 '10 at 21:34

This doesn't quite answer your questions about (2), but one could say the following: given a metric, there is an underlying uniform structure. We can then form the completion with respect to this uniform structure, which is universal for uniformly continuous maps into complete uniform spaces. That this map is injective is due to the existence of "enough" complete uniform spaces. (And to show this, one constructs the completion! I am still thinking about alternative, less constructive, approaches, to this.)

But I think now one can relate (2) to (1) via the following lemma:

If $X$ is a uniform space, and $d$ is a metric on $X$ inducing the given uniform structure, then $d$ extends uniquely to the completion of $X$.

Proof: Something along the lines of: $d$ is uniformly continuous from $X \times X$ to $\mathbb R$, and so extends.

Thus the universal object for (2) had to also be the universal object for (1).

| cite | improve this answer | |
  • $\begingroup$ This sounds good to me so far, Matt. I'll think about it; thanks. (Is "uniformally" an Australianism?) $\endgroup$ – Pete L. Clark Jan 13 '10 at 4:44
  • $\begingroup$ No, uniformally is a consistent misspelling on my part! (I'll correct it.) $\endgroup$ – Emerton Jan 13 '10 at 4:46
  • 2
    $\begingroup$ +1 For uniform spaces! (They're making a comeback!) $\endgroup$ – Harry Gindi Jan 13 '10 at 10:54
  • $\begingroup$ I accept this answer. The reason for the inordinate delay was that there was some aspect of the question that I still wanted to work out in my own mind (something about how best to think about the completion of a uniform space, perhaps?). But I have long since passed the point in my course where this was needed, so I am unlikely to come back to it very soon. Anyway, this was helpful for my course notes: see math.uga.edu/~pete/8410Chapter2v2.pdf $\endgroup$ – Pete L. Clark Mar 29 '10 at 19:37

The canonical universal property (in so far as it is given as an example in Maclane's book :P ) can be stated: the full subcategory of complete metric spaces is reflective in the category of metric spaces and uniformly continuous maps, and the completion functor is the reflector.

| cite | improve this answer | |
  • $\begingroup$ OK, I can see that I will have to have a look at Mac Lane's book (I assume you mean Categories For the Working Mathematician). But isn't everything you said true with "uniformly continuous map" replaced by "isometric embedding"? $\endgroup$ – Pete L. Clark Jan 13 '10 at 4:40
  • $\begingroup$ Yes, but only taking isometric embeddings results in a weaker universal property. One could ask: if $\mathbf{Met}$ is the category of metric spaces and continuous maps, what subcategories $\mathbf{Met}'$ of $\mathbf{Met}$ containing all objects have the property that their full subcategory spanned by the objects which are complete is reflective? $\endgroup$ – Mariano Suárez-Álvarez Jan 13 '10 at 5:03
  • $\begingroup$ For reference: It's in <i>CftWM</i>, pp. 56-57 in my edition (which agrees with Google Books!) It doesn't seem to be discussed much beyond that. $\endgroup$ – Harrison Brown Jan 17 '10 at 6:22

Dear Pete,

I want to give a proof that if $X$ is a metric space, and if $x_0$ and $x_1$ are two distinct points of $X$, then there is a map $f:X \rightarrow Y$ that is uniformly continuous, with $Y$ complete, and such that $f(x_0) \neq f(x_1)$. The main point is that my proof won't refer to the completion $\widehat{X}$ of $X$. It will then give a proof that $X \rightarrow \widehat{X}$ is necessarily injective, without refering to the construction of $\widehat{X}$.

The construction is simple: take $Y = {\mathbb R}$, and define $f(x) = d(x,x_0).$

(Note: slightly edited from the first version, which had unnecessary complications in the definition of $f$.)

| cite | improve this answer | |
  • $\begingroup$ What you suggest is correct. The point of my answers is to try to explain why the adjunctions (1) and (2) are the same, and (from the point of view of (2)) to explain why the unit is injective without making an explicit construction. I took this to be part of what the question was asking. $\endgroup$ – Emerton Jan 19 '10 at 14:31
  • $\begingroup$ But suppose one didn't know about $i_X$, and wanted to show, without making an explicit construction, that $\tilde{i}_X$ was injective, and in fact could be promoted to being an isometric embedding. How would one do this? I think this was one aspect of the quesion, and in any case, is what my answers were intended to explain. $\endgroup$ – Emerton Jan 19 '10 at 15:11
  • $\begingroup$ [deleted previous comments] $\endgroup$ – user2734 May 17 '10 at 21:34
  • $\begingroup$ In case I never thanked you for this additional answer...thanks. $\endgroup$ – Pete L. Clark Sep 5 '13 at 22:10
  • $\begingroup$ @PeteL.Clark : Dear Pete, You're welcome. Cheers, Matt $\endgroup$ – Emerton Sep 6 '13 at 5:13

Mike Shulman give the impression here that understanding the completion of a uniform space is a little trickier than for a metric space:

Cauchy completion of a metric space is, of course, an instance of Cauchy completion of enriched categories. I believe that Cauchy completion of a uniform space is actually also an instance of a general categorical notion of Cauchy completion, but in the more general setting of an equipment (namely, the equipment of sets and filters). See "Categorical interpretation" at uniform space for a too-brief summary of this point of view.

| cite | improve this answer | |
  • $\begingroup$ Perhaps from a certain viewpoint. But from a naive topological viewpoint, I don't think that it's not much different to the jump from limits of sequences to limits of nets/filters, which is to say, non-trivial the first time you see it, but not so much trickier in the end. $\endgroup$ – Emerton Jan 13 '10 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.