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Hey all,

It is well-known that any uniformity is generated by the family of pseudometrics which are uniformly continuous from the product uniformity to $\mathbb{R}$. Further, the uniformity is generated by a metric iff it has a countable base, and in this case it is (Hausdorff) complete iff the metric is complete.

Otherwise, can we say anything about the metrics that generate the uniformity? Specifically, for all complete uniformities, is some sub-uniformity generated by a complete metric?

I would like some way to characterize completeness in terms of the generating metrics.

Edit: A more precise way to state this is: does there exist a property (P) such that a uniformity is complete iff the set of metrics generating the uniformity has property (P).

Here is a nontrivial counterexample to the case where (P) is "some metric is complete", or even "every metric in a set of metrics sufficient to generate the uniformity is complete":

Consider $C[0,1]$ and the pseudometrics $d_x(f,g) = |f(x) - g(x)|$ for each $x\in[0,1]$. These are all complete, but the uniformity generated by them all is the pointwise-convergence uniformity, which is not.

However, not all is lost. Consider now only those $d_x$'s for which $x$ is rational. The subuniformity generated by these has a countable base and is therefore metrizable, i.e., is generated by some metric $d$. This uniformity is still incomplete because the tails of $\lbrace x^n \rbrace$ still form a noncovergent Cauchy sequence, so $d$ is incomplete.

So here is my new conjecture: A uniformity is complete iff every subuniformity with a countable base is complete. Equivalently, (P) is "every metric is complete".

Edit 2: This also does not work, because of the example I gave in the comments, of $d_1(x,y) = |x - y|$ and $d_2(x,y) = |1/x - 1/y|$. Each of these is incomplete on $\mathbb{R}\setminus\lbrace0\rbrace$, but together they form a complete conformity.

Edit 3: Peter Michor suggested I look at the ordering of the generated pseudometrics. Specifically, define a system of pseudometrics as a set for which every pair has an upper bound; then perhaps (P) is "there exists a subsystem of pseudometrics for which every pseudometric is bounded above by something in the subsystem".

However, consider the uniformity of pointwise convergence on $[0,1]$, and this uniformity except with the zero function deleted. The former is complete, the latter incomplete, but the pseudometrics which generate them coincide everywhere they are all defined. In particular, as posets they will be identical. So we see that we cannot determine completeness from the ordering of the pseudometrics.

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  • $\begingroup$ I do not quite understand your question, I am sorry. Are you asking if something like "A uniformity is complete if and only if the family of pseudometrics satisfies a certain property (P)" holds, without assuming that the uniformity is generated by a single metric? $\endgroup$
    – Filippo Alberto Edoardo
    Commented Jun 10, 2013 at 3:35
  • $\begingroup$ Yes, or more generally "A uniformly U is complete if and only if -some- family of pseudometrics generating U satisfies a certain property (P)". If the property (P) is "at least one pseudometric is complete", then I suspect that this is true, though I'm having trouble (dis)proving it. $\endgroup$
    – Andrew Poelstra
    Commented Jun 10, 2013 at 4:00

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A locally convex vector space is complete iff every Cauchy net converges: It has to be Cauchy in each seminorm; and then has to converge in each each seminorm. So I guess you better look at: Property (P) Each pseudometric is complete.

Edit: In locally convex spaces one usually has directed systems of seminorms. So consider a system of pseudo metrics where for any two there is one which is larger then either one. And then the property could be: There is subsystem of complete pseudometrics which is cofinal, so that for any there is one in the subsystem which is larger.

This is just a proposal --- I did not check it.

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  • $\begingroup$ Well, here is a simple example which made me doubt this: consider $\mathbb{R}\setminus\{0\}$ with the two metrics $d_1(x,y) = |x-y|$ and $d_2(x,y) = |1/x - 1/y|$, which are both Euclidean metrics except I've permuted the line for the second one. Neither is complete, since the sequence $\{1/n\}_{n\in\mathbb{N}}$ fails to converge for $d_1$ and the sequence $\{n\}_{n\in\mathbb{N}}$ fails for $d_2$, but the uniformity generated by them is (since it could just as well be generated by the complete metric $\max(d_1,d_2)$). $\endgroup$
    – Andrew Poelstra
    Commented Jun 10, 2013 at 15:42
  • $\begingroup$ Thanks Peter. I will spend some time thinking about your "cofinal subsystem" suggestion and edit the main post if I can find a counterexample (or prove that it works). $\endgroup$
    – Andrew Poelstra
    Commented Jun 21, 2013 at 17:40
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Considering locally convex spaces as suggested by Peter Michor hints at typical counterexamples because the closed graph theorem usually forbids coarser Frechet space topologies. A concrete example is the complete LF-space $\mathscr D$ of test functions for distribution theory (smooth functions on $\mathbb R$ with compact support) for which the closed graph theorem holds.

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If I understand your question correctly (looking at the answers above I seem to detect some uncertainty), then an example of a complete uniform space on which there are NO continous metrics would be a counterexample. Take an uncountable product of, say, closed unit intervals. Any continuous function thereon depends on at most countable coordinates and so this space has the required property.

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