If $2^{\kappa}<\lambda$, how many subsets of size $\kappa$ are there of a set of size $\lambda$.

Question

Assume both cardinals are infinite. Also assume AC as needed.

So, the obvious bound is that there are no more than $\lambda^\kappa\leq 2^\lambda$ of them. But it seems there should be an easy bound smaller than this. I'm especially interested in seeing proofs of any better bound that are nice and short, even if they assume extra whirligigs.

Edit: By ``extra whirligigs'' I mean that I am willing to assume extra relationships between $\kappa$ and $\lambda$. I would prefer, as far as possible, to find an answer with $\lambda$ an arbitrary infinite cardinal. On the other hand, I'm really not too picky, and am as interested in seeing the argument techniques as getting "the best possible" bound.

Answer 1

Some basic observation:

1. If $\kappa\geq\lambda$ then $\lambda^\kappa=2^\kappa$, in which case $2^\lambda$ may be smaller or equal than the result. So we may assume from here on end that $\kappa<\lambda$.

2. It is consistent that $\kappa<\lambda\leq2^\kappa\leq2^\lambda$. However it is consistent that all the inequalities are sharp and $\lambda^\kappa=2^\kappa$. For example, consider the case where $2^{\aleph_0}=\aleph_2$ and $2^{\aleph_1}=\aleph_3$. Now take $\kappa=\aleph_0$ and $\lambda=\aleph_1$. The result is that $\lambda^\kappa=\aleph_2>\lambda$.

3. Even if we require that $2^\kappa<\lambda$ it is still possible that $\lambda<\lambda^\kappa$. If $\lambda$ is singular and its cofinality is $\leq\kappa$ then $\lambda^\kappa>\lambda$. In fact, in that case if the gap is larger than $1$ (i.e. $\lambda^\kappa>\lambda^+$) then we even have that $(\lambda^+)^\kappa>\lambda^+$ and that is a regular cardinal.

   This case is consistent, although now we need large cardinals to prove it.

4. It is also possible that $\kappa$ is sufficiently small (i.e. $\kappa<2^\kappa<\operatorname{cf}(\lambda)\leq\lambda$) and $\lambda^\kappa=\lambda<2^\lambda$. Of course under the same assumptions we can have $\lambda^\kappa=2^\lambda$, e.g. as in the case with $\lambda^+$ in the previous possible scenario.

So all in all there isn't much we can say. Cardinal exponentiation is a wildcard in set theory. There are ways to control it, but we don't consider the actual exponents, but rather all sort of "strangely defined objects" about subsets of $\lambda$ of size $\kappa$ and so on. To learn more about this you may want to read up on PCF theory which was developed in order to prove certain results of this flavor.