Timeline for "All retracts are closed" as separation axiom
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 28, 2022 at 17:45 | comment | added | Steven Clontz | A couple other notes for posterity: local compactness need only fail at at least one point to break T2 in the compactification, not precisely. The proof that compactly generated KC spaces produce KC compactifications in general is Thm 5 of jstor.org/stable/2316017 We investigated some of this at mathoverflow.net/questions/434451 | |
| Nov 27, 2022 at 16:43 | comment | added | Paul Fabel | To see why we must also assume $X$ is compactly generated, let $X$ be the plane, but refine the usual topology so that each countable set is closed. Then no infinite subset of $X$ is compact. However if $Y$ is the Alexandroff compactifictation then each subset of $Y$ which contains $\infty$ is compact. In particular $Y \setminus 0$ is compact but not closed in $Y$. | |
| Nov 27, 2022 at 16:18 | history | edited | Paul Fabel | CC BY-SA 4.0 |
Modifed the original answer, added the extra assumption that $X$ is compactly generated.
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| Nov 27, 2022 at 16:16 | comment | added | Paul Fabel | Thanks Steven, nice catch! We should also assume $X$ is compactly generated, i.e. $A$ is closed in $X$ iff $A \cap C$ is closed in $X$ for each compact $C \subset X$. | |
| Nov 26, 2022 at 17:47 | comment | added | Steven Clontz | How do you know each compact subspace is closed without the Hausdorff property? | |
| Dec 18, 2014 at 12:20 | vote | accept | Dominic van der Zypen | ||
| Dec 18, 2014 at 11:53 | history | answered | Paul Fabel | CC BY-SA 3.0 |