Timeline for Injectivity of cardinality of power set
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 6 at 8:09 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 26, 2011 at 16:18 | vote | accept | Jun Zhang | ||
| Jun 11, 2011 at 1:26 | comment | added | Gerhard Paseman | I can't think of any more gold to add. Thanks for the additional comments. Gerhard "Ask Me About System Design" Paseman, 2011.06.10 | |
| Jun 11, 2011 at 0:44 | comment | added | Joel David Hamkins | And Gerhard, you are correct that it is because of Goedel's work that we know ZFC+GCH is consistent, which I assumed implicitly in my answer. | |
| Jun 11, 2011 at 0:42 | comment | added | Joel David Hamkins | Yes, it is also independent of ZFC+CH. Much of the full answer to this comes from Eason's theorem, which allows you to control completely the continuum function on the regular cardinals. For example, you can have $2^\omega=\omega_1$, whilst $2^{\omega_1}=2^{\omega_2}=\omega_3$, with GCH elsewhere. Also, you can have $2^{\aleph_n}=\aleph_{n+2}$ for $n\gt 0$, but $2^\omega=\aleph_1$, and GCH elsehwere, which satisfies injectivity, plus CH, but not GCH. | |
| Jun 11, 2011 at 0:27 | comment | added | Gerhard Paseman | Your exemplary explanation (at this writing) does not mention whether the poster's statement is independent of ZFC + CH, and your explanation might benefit from mentioning Goedel's work on the consistency of GCH with respect to ZF and ZFC. Gerhard "Gilded Lilies Are Pretty Too" Paseman, 2011.06.10 | |
| Jun 11, 2011 at 0:07 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 421 characters in body
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| Jun 10, 2011 at 23:07 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Improved exposition, added links
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| Jun 10, 2011 at 22:39 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 430 characters in body; added 103 characters in body
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| Jun 10, 2011 at 22:23 | comment | added | Andrés E. Caicedo | Additionally, one can arrange by class forcing that for any $\kappa$ there is a $\lambda\ne\kappa$ with $2^\kappa=2^\lambda$. | |
| Jun 10, 2011 at 22:06 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |