# Cardinal exponentiations inequality

Let $$\kappa < \beth_2$$ and $$\lambda<\beth_1$$ be cardinals. What can we say about $$\kappa^{\lambda}$$ without assuming CH? Is it true that $$\kappa^{\lambda} < \beth_2$$ or $$\kappa^{<\beth_1} < \beth_2$$?

• The title was changed to something I disagree with, because "exponent" refers to a quantity denoted by a superscript, and we are not concerned with inequalities between these, but rather with values of exponentiation as a binary operation on cardinals. Thus, I think the original title was more correct than the so-called "improvement" by user64494 -- even if the plural of "exponentiation" sounds awkward to some ears. – Todd Trimble Jun 7 at 18:39
• @ToddTrimble I agree and have rolled back. – Noah Schweber Jun 7 at 20:44

Even if CH holds, this can break.

Suppose $$\beth_1=\aleph_1$$ and $$\beth_2=\aleph_{\omega+1}$$. Let $$\lambda=\aleph_0$$ and $$\kappa=\aleph_\omega$$. By Konig's theorem we know that $$\kappa^\lambda=\kappa^{cf(\kappa)}>\kappa$$; since $$\kappa^+=\beth_2$$ this means $$\kappa^\lambda\ge\beth_2$$ (which in turn means $$\kappa^\lambda=\beth_2$$, of course).

• Thank you, I think I understand. – dusan Jun 7 at 16:04