Why continuum function isn't strictly increasing?
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?
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:
- $\kappa<\lambda\implies F(\kappa)\leq F(\lambda)$,
- $\operatorname{cf}(\kappa)<\operatorname{cf}(F(\kappa))$
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.)