Cardinality of $A^B$ when $A > B \ge \aleph_0$

Question

For an infinite cardinal A, then

1. if $B$ is finite $A^B = A$
2. If $B$ is infinite and $B \ge A$ then $A^B \ge A^A \ge 2^A > A$
3. What if $B$ is infinite, but $B < A$, i.e. $A > B \ge \aleph_0$. Is there a relationship between $A^B$ and $A$ in this case ?

Answer 1

We can't quite always say. The continuum hypothesis is equivalent to $\aleph_1^{\aleph_0}=\aleph_1$. But it is not provable, so this equality can hold or fail. But $\aleph_1\leq\aleph_1^{\aleph_0}$ is provable. Therefore, $$\lnot\sf CH\iff\aleph_1<\aleph_1^{\aleph_0}.$$

So the only thing you can really say is that $A\leq A^B$. Nothing much more.

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You can also find more intricate problems.

It is always true that $\kappa^{\operatorname{cf}(\kappa)}>\kappa$. So for example, we know that $\aleph_\omega^{\aleph_0}>\aleph_\omega$. But by how much? Is it $\aleph_{\omega+1}$? Is it $\aleph_{\omega+2}$? Is it $2^{\aleph_\omega}$? We can't say, even if we assume that $2^{\aleph_n}<\aleph_\omega$ for all $n<\omega$.

Even worse, what about $\aleph_{\omega_1}^{\aleph_0}$? As far as provability goes, it has no obligation being larger, but it can and it can stay the same (not simultaneously, of course).

But look at $\beth_{\omega_1}$. We can prove that $\beth_{\omega_1}^{\aleph_0}=\beth_{\omega_1}$. So you can't prove exponentiation always grows, and you can prove that it can grow in the case of singular cardinals.