Timeline for compact quotient
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2012 at 9:37 | vote | accept | André Henriques | ||
| Aug 6, 2012 at 18:27 | history | edited | Nik Weaver | CC BY-SA 3.0 |
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| Aug 5, 2012 at 18:28 | comment | added | BS. | You're again right. Too bad... | |
| Aug 5, 2012 at 18:00 | comment | added | Nik Weaver | I don't quite see how being sequential suffices for my argument ... | |
| Aug 5, 2012 at 17:22 | comment | added | BS. | You're right. On the other hand, since $X$ is first countable, its quotient is sequential (the topology is determined by convergent sequences), and this suffices for your argument. This implies in turn that the quotient is a quotient of a compact sec. count., hence metrisable space (by your result). Finally this implies the quotient is a subspace of the Hausdorff compact metrisable space of compacts of $X$ (the equivalences classes), hence is metrisable, i.e. second countable. Thus compactness of the quotient helps, eventually. | |
| Aug 5, 2012 at 15:29 | comment | added | Nik Weaver | @BS: a quotient of a second countable space need not be second countable, or even first countable. Let $X$ be the disjoint union of countably many copies of the one point compactification of ${\bf N}$, and let $\sim$ be the equivalence relation which identifies all the copies of $\infty$ ... but does assuming $X/\sim$ is compact help? | |
| Aug 5, 2012 at 15:01 | comment | added | BS. | You don't need to assume first countability of the quotient space, since it is automatically second countable. | |
| Aug 5, 2012 at 12:04 | history | edited | Nik Weaver | CC BY-SA 3.0 |
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| Aug 5, 2012 at 11:56 | history | answered | Nik Weaver | CC BY-SA 3.0 |