Timeline for Cardinals in $ZFC+\neg CH$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2019 at 0:21 | vote | accept | Jörg Neunhäuserer | ||
| Apr 28, 2019 at 0:11 | comment | added | Andreas Blass | The answer to the added, more precise question is yes: If ZFC is consistent, it remains so when one adds that the set in the question is uncountable. | |
| Apr 27, 2019 at 22:03 | comment | added | Noah Schweber | @JörgNeunhäuserer Much of the general situation - much broader than merely $\mathbb{R}$ - is summarized by Easton's theorem. This isn't the end of the story, but in some sense it shows that many (if not most) of the "naive" questions about cardinality can't be resolved in ZFC alone. | |
| Apr 27, 2019 at 20:12 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
added 1 character in body
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| Apr 27, 2019 at 20:10 | history | edited | Jörg Neunhäuserer | CC BY-SA 4.0 |
added 185 characters in body
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| Apr 27, 2019 at 19:53 | answer | added | Andreas Blass | timeline score: 10 | |
| Apr 27, 2019 at 19:49 | comment | added | Jörg Neunhäuserer | Dear Monroe Eskew, where do i find such results? | |
| Apr 27, 2019 at 19:36 | answer | added | Gerhard Paseman | timeline score: 0 | |
| Apr 27, 2019 at 18:41 | comment | added | Asaf Karagila♦ | If I told you that I can prove that $A\subseteq\Bbb R$ is either empty or it's not, and then I say "now assume it's not empty". What is the cardinality of $A$? | |
| Apr 27, 2019 at 18:14 | comment | added | Monroe Eskew | It is independent. It can take nearly any ordinal value. | |
| Apr 27, 2019 at 18:10 | history | asked | Jörg Neunhäuserer | CC BY-SA 4.0 |