# The definition of cardinal exponentiation

In ZF, Is provable that the set of all functions from B to A can be well ordered for all well ordered sets A and B?

If it holds, for some well ordered set X, 2 to the X is well orderable. Therefore the power set of X is also well orderable, but it couldnt be possible because it implies the axiom of choice. so it does not hold and we also cannot mention that for every cardinal a and b, a to the b is a cardinal. Is this argument right? Im really confused. (The definition of cardinal of X, where X is well orderable, is the unique initial ordinal equipotent to X)

• Since you already know the second paragraph, you can already translate your question to "Does ZF prove AC?" and the answer there is a well known "no". Apr 27 '20 at 6:56

No, there are models of $$ZF$$ where $$2^\omega$$ cannot be well-ordered.