Timeline for Smallest $\beta$ such that it is provable that $2^{\aleph_\beta} > 2^{\aleph_0}$
Current License: CC BY-SA 4.0
4 events
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| Jan 17, 2019 at 16:36 | comment | added | Andrés E. Caicedo | @Dominic Yes, indeed, we may have $2^{\aleph_0}$ to be a fixed point of the sequence of alephs, that is, a $\kappa$ such that $\kappa=\aleph_\kappa$ so that for any $\beta<2^{\aleph_0}$, also $\aleph_\beta<2^{\aleph_0}$, and simultaneously we may arrange that $2^\lambda= 2^{\aleph_0}$ for all infinite $\lambda<2^{\aleph_0}$. | |
| Jan 17, 2019 at 14:40 | comment | added | Dominic van der Zypen | Thanks for your lovely answer @andresecaicedo! - So I take it that it is consistent that $\aleph_\beta \leq 2^{\aleph_0}$ for all $\beta\in 2^{\aleph_0}$? In that case, the last sentence of your post provides the perfect answer to my question. | |
| Jan 17, 2019 at 12:29 | vote | accept | Dominic van der Zypen | ||
| Jan 16, 2019 at 18:27 | history | answered | Andrés E. Caicedo | CC BY-SA 4.0 |