Timeline for The product of Lindelöf spaces
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2019 at 13:15 | comment | added | Jiachen Yuan | Lemme give a guess, first it looks like that being Lindelof is a property which is equivalent to being a stationary set which reflected at every point of uncoutable cofinality. The product topology is equvalent to the topology generated by triangles which allows you to make an alignment via Godel function. But the Godel function is closed. Therefore I think is possible to show that X is Lindelof. | |
| Jun 3, 2019 at 17:07 | comment | added | Anonymous | Isn't it the case that if $X$ consists of all isolated points of $\omega_1 + 1$ along with the last point, then $X$ is Lindel\"{o}f? | |
| Jun 3, 2019 at 9:25 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
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| Jun 2, 2019 at 17:50 | comment | added | Joel David Hamkins | Is it true that a set of ordinals is Lindelöf in the order topology if and only if it contains all its limit points, except for countably many points of countable cofinality? | |
| Jun 2, 2019 at 16:32 | comment | added | Joel David Hamkins | "Countable cofinality" does not tell the whole story. For a set of ordinals to be Lindelöf, it is necessary that the set contain all its limit points of uncountable cofinality. But that is not sufficient, since if it omits uncountably many limit points (of countable cofinality) on a scattered set, then we can also make a violation of the Lindelöf property. | |
| Jun 2, 2019 at 12:43 | comment | added | Péter Komjáth | Wouldn't that imply that the product of countably many copies of $\omega$ is Lindelof? | |
| Jun 2, 2019 at 11:35 | history | edited | user1 | CC BY-SA 4.0 |
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| Jun 2, 2019 at 11:35 | comment | added | user1 | Yes, $X_n$ may not be countable, it only has countable cofinality. | |
| Jun 2, 2019 at 10:34 | comment | added | Santi Spadaro | Which topology does each $\aleph_n$ have? If it's the order topology, then your claim that each $X_n$ must be countable is wrong. If it's the discrete topology then the answer to your question is easily yes, because $X$ would then be homeomorphic to the irrational numbers. | |
| Jun 2, 2019 at 9:49 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Jun 2, 2019 at 9:26 | history | asked | user1 | CC BY-SA 4.0 |