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?

  • $\begingroup$ I just noticed the exact wording of the question was not answered by my answer previously. I added more. $\endgroup$ – Asaf Karagila Nov 28 '12 at 2:02
  • $\begingroup$ Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :) $\endgroup$ – Metta World Peace Nov 28 '12 at 2:10
  • $\begingroup$ 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.) $\endgroup$ – Asaf Karagila Nov 28 '12 at 2:11
  • $\begingroup$ 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。 $\endgroup$ – Metta World Peace Nov 28 '12 at 5:59
  • $\begingroup$ See also math.stackexchange.com/questions/74477/… or math.stackexchange.com/questions/376509/… and other questions linked there. $\endgroup$ – Martin Sleziak Jul 4 '14 at 11:41

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.)

  • $\begingroup$ Thank you. It sounds fairly interesting. $\endgroup$ – Metta World Peace Nov 26 '12 at 10:27
  • $\begingroup$ 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). $\endgroup$ – Asaf Karagila Nov 26 '12 at 10:33
  • $\begingroup$ Good advise. Although it is probably too advanced for me, I'll have a look. $\endgroup$ – Metta World Peace Nov 26 '12 at 10:39
  • $\begingroup$ Now that I am by a computer again, I added a bit. $\endgroup$ – Asaf Karagila Nov 26 '12 at 11:39
  • 1
    $\begingroup$ @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.) $\endgroup$ – Andrés E. Caicedo Jul 5 '14 at 14:41

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