Timeline for A problem of non-emptiness of intersections of certain chains of regular open sets
Current License: CC BY-SA 4.0
12 events
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Aug 26 at 6:40 | history | edited | Jukka Kohonen |
tag fix (order lattices)
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Jan 10, 2023 at 0:27 | vote | accept | Rafał Gruszczyński | ||
Jan 4, 2020 at 20:54 | answer | added | KP Hart | timeline score: 5 | |
Dec 23, 2019 at 19:23 | history | edited | Rafał Gruszczyński | CC BY-SA 4.0 |
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Dec 23, 2019 at 14:10 | history | edited | Rafał Gruszczyński | CC BY-SA 4.0 |
added 99 characters in body; deleted 5 characters in body; deleted 2 characters in body
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Dec 23, 2019 at 13:27 | comment | added | Rafał Gruszczyński | Maybe it woube more reasonable to say $\mathcal C$ covers $\mathcal D$ and reserve the coinitiality for mutual covering? | |
Dec 23, 2019 at 13:21 | comment | added | Rafał Gruszczyński | My use of coinitial is the same as this one: en.m.wiktionary.org/wiki/coinitial To be honest I have never encountered a different notion, but this may be my ignorance of course. | |
Dec 23, 2019 at 13:18 | comment | added | Rafał Gruszczyński | Yes, according to my definition one of the chains may happen to converge a larger (smaller) intersection than the other. And one-element chains are excluded. By the definition and by dependent choices every such chain is infinite. | |
Dec 23, 2019 at 12:56 | comment | added | Joel David Hamkins | Your definition of strictly descending seems to rule out a chain with only one element. | |
Dec 23, 2019 at 12:54 | comment | added | Joel David Hamkins | Your definition of co-initial seems strange to me, since it would be that $\mathcal{C}$ is a chain converging to a much smaller intersection than $\mathcal{D}$, but this isn't usually what is meant by "co-initial". Usually, co-initial would imply that the limit is the same. | |
Dec 23, 2019 at 12:10 | history | edited | Rafał Gruszczyński | CC BY-SA 4.0 |
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Dec 23, 2019 at 11:56 | history | asked | Rafał Gruszczyński | CC BY-SA 4.0 |