## Does a uniform space have a closed embedding in a product of metric spaces?

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I am assuming that uniform spaces are Hausdorff (although it probably doesn't matter for this question). It is more-or-less obvious that a uniform space can be embedded in a product of metric space (if d is a semi-metric on the space) form the quotient gotten by identifying pairs of points at d-distance 0 and then map into the product of these quotients), but I would be interested either to know that every uniform space can be embedded as a closed subspace of such a product or an example of one that cannot.

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Having some trouble parsing the question. Likely because of a () mismatch. And do you mean semi-metric or pseudo-metric? As defined here, for instance: en.wikipedia.org/wiki/Uniformity_%28topology%29 – Igor Khavkine Dec 9 at 17:39

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Sorry, I got mixed up between semi- and pseudo-metric. Having just googled them, I see I should have said pseudo-metric. You are right, the question was senseless. But I still would like to know the answer. Maybe I should mention the context. I can classify the limit closure, in the category of uniform spaces, of the metric spaces as those uniform spaces that are closed subspaces of a product of metric spaces and want to know if that is all uniform spaces. If only John Isbell were still around to ask. – Michael Barr Dec 9 at 18:06
Is the notion of "realcompact" what you are looking for? – Todd Eisworth Dec 9 at 18:14
The condition of the question is equivalent to the uniform space being complete. The other concepts mentioned here are topological and so not appropriate in this context. One could, of course, pose the same question for completely regular spaces with, e.g., the finest uniformity compatible with the topology and so obtain results of this type. – jbc Dec 9 at 18:59

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No the answer is not "complete". It is clear that complete uniform spaces are the limit closure of the complete metric spaces, a question I had already answered. As for "realcompact" the uniform spaces whose associated topology is realcompact appear to be the limit closure of the full subcategory whose only object is the reals. I have not yet written up the details of that, so I am not quite certain. – Michael Barr Dec 9 at 20:51
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## 1 Answer

You are looking for the notion of Dieudonne complete spaces (which turn out to be exactly the closed subspaces of products of metric spaces). As Todd mentioned in his comment, this notion is closely related with the notion of realcompactness (the two notions coincide if there are no measurable cardinals). A way to find examples is to look for pseudo-compact non-compact spaces (for instance $\omega_1$ with (a uniformity compatible with) the order topology).

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Thank you.  This is the full answer I was looking for. – Michael  Barr Dec 9 at 20:54
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