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Is there any example that for cardinal numbers $\kappa < \lambda$, we have $2^\kappa = 2^\lambda$?

My guess is that it only depends on whether GCH holds. Is it true?

Metta World Peace
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  • I just noticed the exact wording of the question was not answered by my answer previously. I added more. – Asaf Karagila Nov 28 '12 at 02:02
  • Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :) – Metta World Peace Nov 28 '12 at 02:10
  • I do (and I am). I mean, seriously. 4am. I am going to sleep! (I have to give a talk tomorrow afternoon, and I won't have time to doze off around noon.) – Asaf Karagila Nov 28 '12 at 02:11
  • What should I say?! I don't want to sound like Cee lo in a talent show. But what you have been doing do amaze and impress me。 – Metta World Peace Nov 28 '12 at 05:59
  • See also http://math.stackexchange.com/questions/74477/does-2x-cong-2y-imply-x-cong-y-without-assuming-the-axiom-of-choice or http://math.stackexchange.com/questions/376509/bijection-between-power-sets-of-sets-implies-bijection-between-sets and other questions linked there. – Martin Sleziak Jul 04 '14 at 11:41

1 Answers1

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This is independent of ZFC. It is consistent that there are no such cardinals, for example if GCH holds. Note that $\lambda\leq\kappa\implies2^\lambda\leq2^\kappa$, so it is enough to show that the continuum function is injective.

However it is consistent that $2^{\aleph_0}=2^{\aleph_1}=\aleph_3$.

There is not much we can say about the continuum function in ZFC. This is a dire consequence from Easton's theorem.

Easton theorem tells us that if $F$ is a function whose domain is the regular cardinals and:

  1. $\kappa<\lambda\implies F(\kappa)\leq F(\lambda)$,
  2. $\operatorname{cf}(\kappa)<\operatorname{cf}(F(\kappa))$

Then there is a forcing extension which does not collapse cardinals and for every regular $\kappa$, $2^\kappa=F(\kappa)$ in the extension.

Assume GCH holds and take the function $F(\kappa)=\kappa^{++}$. We can show that in the extension where $F$ describes the continuum function we have $2^{\kappa}=\kappa^{++}$ for regular cardinals, and $F(\mu)=\mu^+$ for singular $\mu$. This means that GCH fails for all regular cardinals, but $2^\lambda=2^\kappa\iff\lambda=\kappa$. So the injectivity of the continuum function holds, while GCH fails.

(If one is not in the mood for a class-forcing, which can be a bit complicated, one can simply start with GCH and set $2^{\aleph_n}=\aleph_{n+2}$ for $n<\omega$, and GCH to hold otherwise instead.)

Martin Sleziak
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Asaf Karagila
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  • Thank you. It sounds fairly interesting. – Metta World Peace Nov 26 '12 at 10:27
  • You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief). – Asaf Karagila Nov 26 '12 at 10:33
  • Good advise. Although it is probably too advanced for me, I'll have a look. – Metta World Peace Nov 26 '12 at 10:39
  • Now that I am by a computer again, I added a bit. – Asaf Karagila Nov 26 '12 at 11:39
  • "There is not much we can say about the continuum function in ZFC." Asaf... – Andrés E. Caicedo Jul 05 '14 at 06:31
  • @Andres: Is there? – Asaf Karagila Jul 05 '14 at 06:33
  • Yes! Of course! First and foremost, there is pcf theory. But even classically: First of all, Easton's is specifically a result about regular cardinals, and extensions of models of $\mathsf{GCH}$. I do not know of an appropriate "Easton's theorem" where the ground model is arbitrary (there are restrictions!). For singulars, on the other hand, we have Silver's theorem, and its extension by Galvin-Hajnal, and the extensions of their results by Shelah (all this is pre-pcf theory, but of course a strong motivation). And not everything is possible on regulars, in the presence of large cardinals. – Andrés E. Caicedo Jul 05 '14 at 06:40
  • @Andres: Yes, we can indeed prove there are certain limitations on the continuum function. And PCF does tell us a bit about how the continuum function behaves at singulars. But it doesn't really tell us what sort of values we have to have. We can only prove certain properties of the continuum function, we can't really decide anything else. Everything else is "If such and such, then such and such", and the line that you quote was written in the context of absolute values, where we can't say much beyond the two properties given. [...] – Asaf Karagila Jul 05 '14 at 09:24
  • [...] I agree that $\sf ZFC$ can tell us a lot about the behavior of the continuum function in terms of consistency and provability from additional assumptions. But it still doesn't tell us a whole lot more than the very basics. Mainly, I think, the issue here is that we interpret "can say about" in two different and incompatible ways. – Asaf Karagila Jul 05 '14 at 09:25
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    @AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $\aleph_\omega$ is strong limit, then $2^{\aleph_\omega}<\aleph_{\omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{\aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.) – Andrés E. Caicedo Jul 05 '14 at 14:41