It is known that the sum and the product of two Dedekind-finite cardinals are also Dedekind-finite cardinals. What about cardinal exponentiation ?
Question: Let A and B be two Dedekind-finite cardinals, let C be the cardinal A power B (i.e:let x be a set with cardinal A and y be a set with cardinal B and let C be the cardinal of the set of functions with domain y and range a subset of x). 
Is it true that C is a Dedekind-finite cardinal ?
Gérard Lang



Exponentiation and Dedekind-finite cardinals
The answer is no, not necessarily, because if there are infinite Dedekind finite sets, then the class of Dedekind finite sets is not closed under power set, and hence not closed under $A\mapsto 2^A$. 


To see this, simply note that if $A$ is any infinite set, then $P(A)$ has the singletons, the doubletons, the subsets of size $3$, and so on. So we can find a countably infinite subset of $P(P(A))$, and so $P(P(A))$ is not Dedekind finite.


In particular, if $A$ is Dedekind finite but infinite, then $2^{2^A}$ is not Dedekind finite, and so it is consistent with ZF that the Dedekind finite sets are not closed under exponentiation.


