Why continuum function isn't strictly increasing?

Question

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?

Answer 1

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.)