Timeline for Ordered union of Borel sets
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Mar 30, 2019 at 5:50 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:49 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:50 | |||||
| Mar 30, 2019 at 5:49 | vote | accept | user137602 | ||
| Mar 30, 2019 at 5:49 | |||||
| Mar 29, 2019 at 7:04 | comment | added | user137602 | Got it. Thank you for the answer! | |
| Mar 28, 2019 at 23:27 | comment | added | Skeeve | @NateEldredge thanks a lot, I've tried to integrate that approach into my answer. | |
| Mar 28, 2019 at 23:25 | history | edited | Skeeve | CC BY-SA 4.0 |
add discussion of cardinality
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| Mar 28, 2019 at 21:52 | comment | added | Skeeve | @bof thanks a lot for the suggestion, I've updated my answer using it. | |
| Mar 28, 2019 at 21:50 | history | edited | Skeeve | CC BY-SA 4.0 |
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| Mar 28, 2019 at 12:51 | comment | added | Nate Eldredge | The comment of Andreas Blass on the question gives another argument that, in ZFC, there exists a non-Borel set of cardinality $\aleph_1$. Again, no CH needed. | |
| Mar 28, 2019 at 12:38 | comment | added | bof | Why don't you just take a non-Borel set of minimum cardinality and well-order it so all initial segments have lower cardinality? That just needs AC, no CH at all. | |
| Mar 28, 2019 at 12:15 | comment | added | Skeeve | Sorry, you are right. It is not clear wether one avoid using weak CH (I will try to come back to this later). | |
| Mar 28, 2019 at 12:13 | history | edited | Skeeve | CC BY-SA 4.0 |
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| Mar 28, 2019 at 12:04 | comment | added | Andrés E. Caicedo | You haven't. You still appear to claim that $2^{\aleph _0}<2^{\aleph_1}$. | |
| Mar 28, 2019 at 11:55 | comment | added | Skeeve | Thanks for pointing this out, I was not aware of these facts! (corrected the answer) | |
| Mar 28, 2019 at 11:54 | history | edited | Skeeve | CC BY-SA 4.0 |
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| Mar 28, 2019 at 11:43 | comment | added | Andrés E. Caicedo | Also, without choice, it is consistent that every set of reals is Borel. | |
| Mar 28, 2019 at 11:10 | comment | added | Goldstern | You seem to use the inequality $2^{\aleph_0}<2^{\aleph_1}$, also known as "weak CH". This is not provable in ZFC. - Also, I don't think you can prove from ZF (without AC) that the reals contain a set of cardinality $\aleph_1$. | |
| Mar 28, 2019 at 9:53 | history | answered | Skeeve | CC BY-SA 4.0 |