Timeline for Is the lexicographic ordering on the unit square perfectly normal?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2020 at 22:55 | answer | added | Henno Brandsma | timeline score: 3 | |
| Jan 28, 2020 at 19:35 | history | became hot network question | |||
| Jan 28, 2020 at 16:58 | vote | accept | VDGG | ||
| Jan 28, 2020 at 16:58 | vote | accept | VDGG | ||
| Jan 28, 2020 at 16:58 | |||||
| Jan 28, 2020 at 16:58 | vote | accept | VDGG | ||
| Jan 28, 2020 at 16:58 | |||||
| Jan 28, 2020 at 16:55 | answer | added | Ramiro de la Vega | timeline score: 5 | |
| Jan 28, 2020 at 16:53 | comment | added | Ramiro de la Vega | @Gro-Tsen, the open unit square would be metrizable (it is just a disjoint union of open intervals). | |
| Jan 28, 2020 at 15:50 | comment | added | Martin Sleziak | It is without proof also in the pi-base. | |
| Jan 28, 2020 at 15:44 | vote | accept | VDGG | ||
| Jan 28, 2020 at 16:58 | |||||
| Jan 28, 2020 at 15:44 | history | edited | Will Brian | CC BY-SA 4.0 |
added 173 characters in body
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| Jan 28, 2020 at 15:41 | comment | added | Will Brian | Thanks @JoelDavidHamkins. The wikipedia entry uses the phrase "precisely separated" which I think makes a difference here -- I'll edit the question to make it more clear. | |
| Jan 28, 2020 at 15:38 | comment | added | Joel David Hamkins | He's using the same definition as in Wikipedia: en.wikipedia.org/wiki/Normal_space. | |
| Jan 28, 2020 at 15:27 | answer | added | Will Brian | timeline score: 5 | |
| Jan 28, 2020 at 15:17 | comment | added | Will Brian | I'm not sure about your definition of "perfectly normal". Doesn't Urysohn's lemma say that, in any $T_4$ space, any disjoint closed sets are separated by a continuous function? The definition I know of "perfectly normal" is that every closed set is a $G_\delta$. This is equivalent to "every nonempty closed set can be separated from its complement by a continuous function onto $[0,1]$". Maybe this is the definition/characterization you're thinking of? | |
| Jan 28, 2020 at 15:09 | comment | added | Nate Eldredge | Steen and Seebach, Counterexamples in Topology, 2e, say that it isn't, but the proof is left as an exercise (problem 71, page 209). | |
| Jan 28, 2020 at 15:06 | comment | added | Gro-Tsen | But one can be completely normal without being compact, can't one? Anyway, even if there's a reason, this sort of things should always be clarified. | |
| Jan 28, 2020 at 15:01 | comment | added | Joel David Hamkins | He means the closed unit square, for otherwise it won't be compact. | |
| Jan 28, 2020 at 14:58 | comment | added | Gro-Tsen | You should clarify whether you mean the closed unit square or the open unit square (or either one indifferently, or some other variant). | |
| Jan 28, 2020 at 14:55 | comment | added | Joel David Hamkins | Wikipedia on this space: en.wikipedia.org/wiki/…. (Very interesting question!) | |
| Jan 28, 2020 at 14:37 | history | edited | YCor | CC BY-SA 4.0 |
fixed numerous typos
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| Jan 28, 2020 at 11:29 | history | asked | VDGG | CC BY-SA 4.0 |