Timeline for Does every Lindelof uniform space have a Lindelof completion?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 9, 2012 at 15:56 | vote | accept | Fred Dashiell | ||
| Jan 9, 2012 at 15:55 | answer | added | Fred Dashiell | timeline score: 1 | |
| Jan 9, 2012 at 15:49 | comment | added | Fred Dashiell | Yes, as KP Hart points out, the question is answered by the space S x S, which is a counterexample. | |
| Jan 9, 2012 at 12:32 | comment | added | Fred Dashiell | Maybe it does! Is it true that if D is a dense subset of a complete uniform space X, then X is the completion of D in the relative uniformity inherited from X? Maybe this is obvious, but I did not write it out until your question made me think it through. I was worried that I have no idea what the uniformity on S x S is. I guess all you need is that the embedding D --> X is uniformly continuous, and of course this is obvious. | |
| Jan 9, 2012 at 11:22 | comment | added | KP Hart | So, why doesn't $S\times S$ answer it? Isn't it the completion of the dense subspace of rational points? | |
| Jan 7, 2012 at 13:35 | history | edited | Fred Dashiell | CC BY-SA 3.0 |
add info about example of Sorgenfrey plane
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| Jan 7, 2012 at 5:10 | history | edited | Fred Dashiell | CC BY-SA 3.0 |
clarification: "is" --> "must be"
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| Jan 7, 2012 at 1:43 | history | edited | Fred Dashiell | CC BY-SA 3.0 |
add related question on Dieudonne complete space
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| Jan 6, 2012 at 4:17 | history | edited | Fred Dashiell | CC BY-SA 3.0 |
added reference
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| Jan 3, 2012 at 6:07 | comment | added | Fred Dashiell | Yes, this was my intent in the original question. | |
| Jan 2, 2012 at 22:15 | answer | added | Henno Brandsma | timeline score: 1 | |
| Jan 2, 2012 at 18:27 | comment | added | Henno Brandsma | 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. | |
| Jan 2, 2012 at 18:24 | comment | added | Henno Brandsma | 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. | |
| Jan 2, 2012 at 15:51 | comment | added | Gerald Edgar | Is there a reasonable criterion of which (completely regular Hausdorff) spaces are completely uniformizable? | |
| Jan 1, 2012 at 23:04 | answer | added | Gerald Edgar | timeline score: 0 | |
| Jan 1, 2012 at 22:38 | history | asked | Fred Dashiell | CC BY-SA 3.0 |