Timeline for Is there any quasi-compact space which is not a quotient of any compact Hausdorff space?
Current License: CC BY-SA 4.0
11 events
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| Apr 11, 2019 at 23:25 | comment | added | Ramiro de la Vega | Here math.stackexchange.com/a/2794129/94514 is a proof that the one point compactification of the rationals is not a continuous image of (so it is not a quotient of) a compact Hausdorff space. | |
| Apr 11, 2019 at 22:07 | comment | added | YCor | @NateEldredge I've understood the question with the usual definition of quotient topology. | |
| Apr 11, 2019 at 22:06 | comment | added | Rick Sternbach | It is the usual one, not the identification space. Although examples involving identification space is of course welcome. But this is equivalent to the quotient case, since identification map is just a quotient map onto a closed subspace. | |
| Apr 11, 2019 at 21:40 | comment | added | Nate Eldredge | "Quotient" here has a different definition than the usual one, I guess. It should be a new topology on X, rather than on the set of equivalence classes? | |
| Apr 11, 2019 at 19:46 | comment | added | Rick Sternbach | Any such space would be compactly generated, maybe this gives some hints. | |
| Apr 11, 2019 at 19:44 | comment | added | Rick Sternbach | I also suspect most of affine algebraic varieties with Zariski topology would be such examples, but again I have no proof or disproof of this. | |
| Apr 11, 2019 at 19:05 | history | edited | YCor | CC BY-SA 4.0 |
repeated question inside the text
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| Apr 11, 2019 at 19:04 | comment | added | YCor | What about an infinite countable set with the compact topology whose nonempty open subsets are the cofinite subsets? I don't see at the moment whether it's quotient of a Hausdorff compact space. | |
| Apr 11, 2019 at 19:00 | review | Close votes | |||
| Apr 11, 2019 at 20:10 | |||||
| Apr 11, 2019 at 18:25 | review | Low quality posts | |||
| Apr 11, 2019 at 18:34 | |||||
| Apr 11, 2019 at 18:08 | history | asked | Rick Sternbach | CC BY-SA 4.0 |