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I am learning the set theory with the textbook "Set theory" written by Abhijit Dasgupta. I am confused about the description outlined at page 206. The description says that when the exponent is infinite, then we can't compute the cardinal power such as 2^{\aleph_0} as an aleph even using the AC. My question is that under the AC, since any infinite cardinal is an aleph and 2^{\aleph_0} is infinite, 2^{\aleph_0} should be \aleph_{\alpha} for some ordinal \alpha. What is wrong here?

Asaf Karagila
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Xiaojiang Ye
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1 Answers1

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The emphasis is on the word "compute". You are correct that there is some ordinal $\alpha$ such that $2^{\aleph_0}=\aleph_{\alpha}$, but with ZFC alone we cannot pin down the value of $\alpha$. With forcing, you can find a model of ZFC in which $2^{\aleph_0}=\aleph_1$ (the continuum hypothesis), but also a model with $2^{\aleph_0}=\aleph_{42}$. A more general result ist Easton's theorem.

Martin Brandenburg
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  • Thank you very much. I will learn why $2^{\aleph_0}$ could be different $\aleph_{\alpha}$ under different model following the direction you point out. Thank you again. – Xiaojiang Ye Oct 20 '24 at 06:40
  • Why does $2^{\aleph_0}$ have to be an $\aleph_\alpha$? – MJD Oct 20 '24 at 07:51
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    @MJD $\mathsf{AC}$ is equivalent in $\mathsf{ZF}$ to all sets being well-orderable. – J.G. Oct 20 '24 at 07:52
  • I think the full argument goes like this: $2^{\aleph_0}$ is well-orderable, therefore is order-isomorphic to some ordinal $\beta$. Let $\mathcal B$ be the set of ordinals equinumerous with $\beta$. $\mathcal B$ is nonempty and therefore has a least element $\gamma$. This $\gamma$ is an $\aleph$ number because the $\aleph$s are by definition the least elements of equivalence classes such as $\mathcal B$. Is that right? – MJD Oct 22 '24 at 15:54