Is $\kappa^\lambda=2^\lambda$($2 \le \kappa<\lambda$,$\lambda$ infinite) valid in set models of ZF?

Question

Let $2 \le \kappa<\lambda$(both cardinal numbers), in which $\lambda$ is infinite. Then these formula as follows hold where in ZFC:

1. $\lambda+\kappa=\lambda$

2. $\lambda\cdot\kappa=\lambda$

3. $\kappa^\lambda=2^\lambda$

However, if $\lambda$ is not Dedekind-infinite, then 1,2 fail.

But for 3, it's not quite clear.

To prove it. Obviously $2^\lambda\le\kappa^\lambda$; for the other direction, I only got $\kappa^\lambda\le 2^{\kappa \cdot \lambda}$, but $2^{\kappa \cdot \lambda}=2^{\lambda}$ seems not valid.

So my question: Is 3 also valid in all set models of ZF?

Answer 1

Assuming that $\lambda\cdot\kappa=\lambda$, yes -- as the proof follows through immediately. In particular when the two are ordinals then it is true.

However for general cardinals this may be false. For example if $\lambda$ is the cardinal of an amorphous set then $2^\lambda$ is Dedekind-finite. It follows that $3^\lambda$ is strictly larger than $2^\lambda$, otherwise we could have omitted some of the functions and retain the same cardinality, which would imply that $2^\lambda$ is Dedekind-infinite.