Timeline for A topology on $\Bbb R$ where the compact sets are precisely the countable sets
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Nov 22, 2016 at 20:56 | comment | added | Lajos Soukup | Will Brian's argument used (a weak version of) the Axiom of Choice. Is it possible to prove the non-existence in ZF without using AC? | |
| Nov 21, 2016 at 20:48 | vote | accept | Cauchy | ||
| Nov 21, 2016 at 20:41 | comment | added | Forever Mozart | Although, in $\omega$ with the cofinite topology every subset is compact, but clearly there is a decreasing sequence of sets with empty intersection. So it seems that you need to use the fact that $\mathbb R$ is uncountable... | |
| Nov 21, 2016 at 20:30 | comment | added | Pietro Majer | A decreasing sequence of countable sets may have empty intersection | |
| Nov 21, 2016 at 20:29 | answer | added | Will Brian | timeline score: 40 | |
| Nov 21, 2016 at 20:25 | comment | added | Will Brian | @ForeverMozart: Yes, that's exactly the thing I had in mind. | |
| Nov 21, 2016 at 20:24 | comment | added | Forever Mozart | @WillBrian Because every infinite Hausdorff space contains a countably infinite discrete subspace, right? | |
| Nov 21, 2016 at 20:19 | comment | added | Will Brian | Just to narrow the search a bit, let me point out that if every countable subset of a space is compact, then the space is either finite or it fails to be Hausdorff. | |
| Nov 21, 2016 at 20:14 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
edited for clarity
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| Nov 21, 2016 at 19:42 | review | First posts | |||
| Nov 21, 2016 at 20:14 | |||||
| Nov 21, 2016 at 19:41 | history | asked | Cauchy | CC BY-SA 3.0 |