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Recall Lindelof = every open cover has a countable subcover. Is the question of the title answered by known material from some standard text on uniform spaces?

Note it is well known to be true for metric spaces, since a metric space is separable if and only if it is Lindelof (see e.g., Engelking Gen Top 1989, 4.1.16).

[ADDED] In addition to Howes book (mentioned below), I have looked at a paper by T. Ishii, "Paracompactness and topological completion", Fund Math 92 (1976), 65-77. Theorem 4.1 (p. 74) of that paper seems to come close to the question. I think part (c) ==> (a) of that theorem implies that if X is Lindelof then the finest uniformity compatible with X has a completion which is Lindelof. But the proof is highly indirect and I don't follow all of it yet. And I am also not sure that does what we need: the completion of X in its original uniformity must be Lindelof.

This question deserves an answer, preferably with a direct proof.

[ADDED Jan 6] A related question along these lines: If X is a "Dieudonne complete" space (i.e., a space which admits a complete uniformity: see comment below) which is topologically separable (i.e., has a countable dense subset), must X always be a Lindelof space? An example to consider is X = S x S where S is the Sorgenfrey line. This is known to not be Lindelof. Does it have a complete compatible uniformity? I.e., is it "Dieudonne complete"? If it does, it provides a counterexample to the question.

[Add Jan 7] Yes, S x S is topologically complete ("Dieudonne complete"), because S is realcompact and therefore S x S is realcompact (see Engelking p. 217), and every realcompact space is topologically complete (Engelking p. 464). So X = S x S with a complete uniformity is an example of a complete uniform space with a (countable) dense Lindelof subspace, but X is not Lindelof.

However, this does not exhibit S x S as the completion of some uniform space which is Lindelof in the induced topology. So it does not (as yet) contradict the question of this thread: if X is a Lindelof uniform space must its completion (with respect to the original uniformity) also be Lindelof?

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Is there a reasonable criterion of which (completely regular Hausdorff) spaces are completely uniformizable? –  Gerald Edgar Jan 2 '12 at 15:51
    
This property for (completely regular Hausdorff) spaces is called Dieudonné complete (or topologically complete) and such spaces include all paracompact Hausdorff spaces. A theorem by Tamano (1960) characterizes these spaces as: for every $p$ in $\beta(X) \setminus X$ there is a partition of unity of $X$ such that $p$ is not in the closure (in $\beta(X)$) of the support of $f$ ($X \setminus Z(f)$) for all $f$ in that partition of unity. –  Henno Brandsma Jan 2 '12 at 18:24
    
The previous comment of mine implies that every Lindelöf (regular) space $X$ has a uniformity that induces the topology of $X$ and is already complete. So the original poster probably means that the uniformity on $X$ is fixed and given and the question is for its (essentially unique) completion under that uniformity. –  Henno Brandsma Jan 2 '12 at 18:27
    
Yes, this was my intent in the original question. –  Fred Dashiell Jan 3 '12 at 6:07
    
So, why doesn't $S\times S$ answer it? Isn't it the completion of the dense subspace of rational points? –  KP Hart Jan 9 '12 at 11:22

3 Answers 3

this paper by Howes gives a characterization of uniform spaces with a Lindelöf completion, but the characterization uses the derived uniformity, and notions like preparacompactness. I don't see as yet how it would imply, or refute, your question, but the link might help anyway. Howes did a write up of all the theory of uniform spaces and covering properties in his book "Modern Analysis and Topology" as well, and the result can also be found in its 4th chapter.

[added] It is equivalent for a Lindelöf uniform space $X$ to have a paracompact or a Lindelöf completion $X^{\ast}$, by the classical theorem (5.1.25 in Engelking, General Topology): a paracompact space that has a Lindelöf dense subspace is Lindelöf, and $X$ is dense in $X^{\ast}$.

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Thanks Henno. I am looking at Howes, but I also can't untangle it yet. So the original question remains unanswered here. I would be surprised if it is not easily derived from some standard treatment of uniform spaces. –  Fred Dashiell Jan 3 '12 at 6:23

Let $X = \beta S$ be the Stone-Cech compactification of an uncountable discrete space. The uniform structure for $X$ is what you want, I think. The subset $S$ is not Lindelof, but its completion is $X$. (Actually, the one-point compactification should work, too.)

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Thanks Gerald, but your example is for the converse of my question. I need a Lindelof uniform space with no Lindelof completion, or a reference or proof that none such exists. Without uniformizability, there are examples of non-Lindelof spaces with a dense Lindelof subspace: any separable non-Lindelof space will do, such as SxS where S is the Sorgenfrey line. ... Fred –  Fred Dashiell Jan 2 '12 at 3:44
    
You are right, of course. I will leave this answer, so others won't make the same mistake. –  Gerald Edgar Jan 2 '12 at 15:49
up vote 0 down vote accepted

I am entering an "answer" because don't know any other way to mark the question settled. KP Hart added a comment suggesting the space S x S discussed above is a counterexample. This is correct because whenever D is a dense subset of a complete uniform space X, then X is the completion of D in the inherited uniformity. (The embedding D --> X is uniformly continuous.)

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This is a good example of being blinded by prejudice. I thought the statement would be found true, and this influenced my think to the extent that I did not see the counterexample. –  Fred Dashiell Jan 10 '12 at 6:08

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