# Given sets $X$ and $Y$, are the elements of $Y^X$ "determined" under naive set theory?

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?

• I don't think there is a canonical choice for the elements of $X\times Y$. Often we agree that $\{a\}\times \{b\}=\{\{a\},\{a,b\}\}$, but $\{\{b\},\{a,b\}\}$ would work just as well, and historically other choices can and have been made because it really does not matter much what the elements of $A\times B$ are, what matters is the universal property enjoyed by the projection maps. I think the answer for $Y^X$ will be very similar.
– MJD
Sep 26 at 1:10
• @MJD Oh, yeah, of course—I didn't mean at quite this level of granularity. However one chooses to define ordered pairs, there will be a natural bijection between $X\times Y$ under your definition and $X\times Y$ under some other definition. This is distinct from the axiomatic issues, e.g. with AoC. Sep 26 at 1:13
• Well, you said “everyone will agree, for instance, that given knowledge of X and Y, we know exactly what should be in … X×Y”. And that made me think I didn't understand what you were asking, because I couldn't imagine what you thought everyone would agree on.
– MJD
Sep 26 at 1:17
• The downward Lowenheim-Skolem theorem essentially kils any hope along these lines. Sep 26 at 1:36
• You can sort of see from the axiom of choice itself that $Y^X$ can't be naively determined, since the axiom asserts the existence of a choice function, which is a member of $X^{\mathcal P(X)\backslash\{\emptyset\}}$ satisfying a certain property.
– M W
Sep 26 at 2:12