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?
 A: 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.)
