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The Feferman - Levy model makes $\aleph_1$ singular by a cardinal collapse $\aleph_1 = \aleph_{\omega}^L$. Unless I've got something wrong, the same thing would work to make any well-orderable cardinal $\alpha$ cofinal in its well-ordered cardinal successor. Is that right?

The Feferman -Levy model also makes the continuum a countable union of countable sets. Does that generalize to Beth numbers, in the sense of successive power sets starting with $\omega$? For each finite $n$, are there models where $\beth_{n+1}$ is a union of $\beth_{n}$ many sets each smaller than or the same size as $\beth_{n}$?

Colin McLarty
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  • Are you familiar with the construction of the Feferman-Levy model (preferably with a slightly modern approach of symmetric extensions, rather than $L(G)$ for some generic set $G$)? Have you tried reiterating the construction with different variables? – Asaf Karagila Oct 13 '12 at 15:46
  • If you are not as familiar, http://mathoverflow.net/questions/100717/zf-the-reals-are-the-countable-union-of-countable-sets-consistent/100725#100725 has some information (including the comments). – Asaf Karagila Oct 13 '12 at 15:48
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    Hi Colin. I am not sure I understand the question in the first paragraph. Are you asking to preserve $\alpha$ and $\alpha^+$? ${\rm cof}(\alpha)$ and $\alpha^+$? Only $\alpha^+$? – Andrés E. Caicedo Oct 13 '12 at 21:30
  • Hi, yes there is an ambiguity. What I was vaguely thinking was to preserve α in the sense of not collapsing any cardinals up to and including it, but making it cofinal in the forcing extension's version of α+. But my main interest is in the second paragraph about finite-indexed beths. – Colin McLarty 0 secs ago – Colin McLarty Oct 13 '12 at 23:19
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    I figured. The issue in the first paragraph is that preserving a singular and collapsing its successor requires large cardinals, by Jensen's covering lemma. – Andrés E. Caicedo Oct 13 '12 at 23:21
  • @ Asaf Karagila. Thanks. But this is a spin off from work on $n$-th order arithmetic foundations for homological algebra (which in fact got a good start during a visit to Colombia). My research goal can be met using $L$. I wonder if it can also be met without the apparatus of constructibility, but if the question on beths is not answered in any reference then I will probably not pursue it. – Colin McLarty Oct 14 '12 at 14:32
  • François G. Dorais answered my question about beth numbers in his answer to "Can iterating countable unions give every set?" He cites a work by Moti Gitik showing the consistency with ZF of the statement that every infinite set is a countable union of sets of smaller cardinality. – Colin McLarty Oct 15 '12 at 04:51
  • Colin, your previous comment did not answer either of my questions from the first comment. I should also point out Gitik's model requires immensely large cardinals; whereas the Feferman-Levy model requires none. – Asaf Karagila Oct 15 '12 at 08:33
  • Asaf, that is interesting about large cardinals. Thanks. As to your questions I know the Feferman - Levy proof uses forcing to make all $\aleph_n^L$ countable, and symmetry to keep $\aleph_{\omega}^L$ uncountable, I do not know exactly how this produces the result on the continuum, and because the research that led me to this question does not actually require answering it I do not mean to work on modifying the proof myself. In fact now that I know the answer, I see it will not let me avoid use of constructibility. – Colin McLarty Oct 15 '12 at 12:43
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    There is no actual use of constructibility. We could have equally have collapsed $\beth_n$'s to $\aleph_0$ and used the same process to ensure that $\beth_\omega$ is the new $\aleph_1$. – Asaf Karagila Oct 15 '12 at 17:43
  • I did not say you use constructibility here. I said my project that led me to this question, actually has to use constructibility. I had thought it might not need to, but it does. – Colin McLarty Oct 15 '12 at 17:54
  • Ah. ${}{}{}{}{}$ – Asaf Karagila Oct 15 '12 at 18:06

1 Answers1

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Posting this as an answer at Colin's request. The second paragraph of the question is addressed at this other MO question.

The answer to the question in the first paragraph is delicate, it depends on how much of the ground model we decide to preserve: $\alpha$ and $\alpha^+$? $\rm{cof}(\alpha)$ and $\alpha^+$? Only $\alpha^+$? The issue is that a straight generalization of Feferman-Lévy must fail just based on consistency strength considerations, because Jensen's covering lemma gives us that preserving a singular and collapsing its successor requires large cardinals (even in $\mathsf{ZF}$). For the covering lemma, see:

  • Devlin, Keith I.; Jensen, Ronald B. Marginalia to a theorem of Silver. In $\models$ISILC Logic Conference (Proc. Internat. Summer Inst. and Logic Colloq., Kiel, 1974), G. H. Müller, A. Oberschelp, and K. Potthoff, eds., pp. 115–142. Lecture Notes in Math., Vol. 499, Springer, Berlin, 1975. MR0480036 (58 #235)
  • Mitchell, William J. The covering lemma. In Handbook of set theory. Vols. 1, 2, 3, Kanamori, Foreman, eds., pp. 1497–1594, Springer, Dordrecht, 2010. MR2768697
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Andrés E. Caicedo
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  • The reference in "MO question." is pointing to ams site. – Eran Oct 15 '12 at 18:29
  • Typical. Thanks, it should be fixed now. – Andrés E. Caicedo Oct 15 '12 at 19:11
  • Wait, how did my answer answer the second paragraph? It suggested that there is some model in which countable unions of countable sets have power sets which can be mapped onto sets of ordinals. This need not imply, methinks, that $\beth_{n+1}$ is a small union small sets. Gitik's model should be such example, though. Perhaps you wanted to link Francois' answer referred to by Colin in the comments? – Asaf Karagila Oct 15 '12 at 20:58