Given sets $X$ and $Y$, are the elements of $Y^X$ determined by naive set theory? What I mean is roughly as follows. For some sets, no special care over axioms is needed for mathematicians to agree on what they should contain. Everyone will agree, for instance, that given knowledge of $X$ and $Y$, we know exactly what should be in $X\cup Y$, or $X \times Y$. Naive reasoning alone is sufficient to determine what these sets contain. Perhaps I am wrong about this. But I believe that if some axiomatization of set theory ever made $X\cup Y$ contain "the wrong elements", we would sooner admit that the axiomatization was wrong than say that this alternate form of $X\cup Y$ was an equally valid one. Thus $X\cup Y$ is "naively determined".
This is not true for, e.g., the Cartesian product of an infinite family of sets. The non-emptiness of arbitrary Cartesian products is equivalent to the Axiom of Choice, and there are plenty of mathematicians who are willing to bite the bullet and say that the Cartesian product of certain infinite families might be empty. Thus, the general Cartesian product is not "naively determined".
I hope this definition of "naively determined" is at least a somewhat sensible one.
Anyway, my question is as follows: is the set $Y^X$ naively determined in this way?
