Timeline for A criterion for second countability
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| May 29, 2018 at 8:52 | vote | accept | ABB | ||
| May 29, 2018 at 5:25 | answer | added | Taras Banakh | timeline score: 4 | |
| Nov 19, 2017 at 14:58 | comment | added | Joel David Hamkins | I would suggest instead that the commenters post their answer as an answer, rather than as a comment. | |
| Nov 19, 2017 at 11:37 | comment | added | YCor | @GABB when a question is so easy that it's answered in comments then I think it's reasonable to modify the question to make it more interesting. So I'd rather suggest to do so (and add relevant tags). I don't mean ask a second question. Still, the question can include a discussion reflecting the input from previous comments. | |
| Nov 19, 2017 at 11:33 | comment | added | YCor | @WilleLiou: standard word for "non-countable" is "uncountable" :) | |
| Nov 19, 2017 at 11:29 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
deleted superfluous exclam
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| Nov 19, 2017 at 6:47 | history | edited | ABB |
edited tags
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| Nov 19, 2017 at 6:43 | comment | added | ABB | Probably you are right. So I do now. | |
| Nov 19, 2017 at 6:41 | comment | added | მამუკა ჯიბლაძე | @GABB Are you sure it makes sense to add these tags to the question the way it is now? I believe you should ask a separate question about topological vector spaces | |
| Nov 19, 2017 at 6:39 | history | edited | ABB | CC BY-SA 3.0 |
edited tags
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| Nov 19, 2017 at 6:33 | comment | added | ABB | This question seems to be much more challenging (and mysterious) when, $X$ is considered as a locally convex topological vector space i.e. $\tau$ induced by a family of seminorms on $X$. It would be great to have an answer in this particular case. | |
| Nov 18, 2017 at 5:16 | comment | added | ABB | @WilleLiou It was really interesting. Thank you very much. | |
| Nov 17, 2017 at 20:56 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Fixes typo in the title
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| Nov 17, 2017 at 20:46 | comment | added | Wille Liu | An example: Let $X$ be a non-countable set. We equip $X$ with the topology whose open sets consist of $\emptyset$ and those subsets $U\subseteq X$ with $|X\setminus U| < \infty$. Then any basis generates the same Borel $\sigma$-algebra but $X$ is not second countable. | |
| Nov 17, 2017 at 20:04 | comment | added | Michael Greinecker | @WilleLiou The Borel $\sigma$-algebra is the power-set, but the $\sigma$-algebra generated by the basis of singletons is the countable-co-countable $\sigma$-algebra. | |
| Nov 17, 2017 at 19:51 | comment | added | Wille Liu | What about non-countable discrete sets? | |
| Nov 17, 2017 at 19:32 | history | edited | ABB | CC BY-SA 3.0 |
added 18 characters in body
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| Nov 17, 2017 at 19:20 | history | asked | ABB | CC BY-SA 3.0 |