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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?

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    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.2014-07-04

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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.)

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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.)2014-07-05