TITLE: Prove that the axiom of choice is necessary in order to prove something else.
QUESTION [33 upvotes]: My mathematical background is perhaps a little lacking on this topic, but I've been searching and haven't come up with a satisfactory answer to this question. I have no idea how to approach the problem or if it has been answered.
I have seen numerous cases where it is asserted that "the axiom of choice is necessary" to complete a particular proof but I've yet to see a case where the necessity of the axiom of choice is actually itself proven.
For example consider the following question: Additive function $:\mathbb{R} \rightarrow \mathbb{R}$ that is not linear.
The answers and comments note that you can't construct such a function without the axiom of choice. Then it seems like this would imply, for example, that any function with a closed form
which is additive is necessarily linear. Is this correct? How would one go about proving the necessity of the axiom of choice?
I apologize in advance that I don't have my own attempt to share since I'm not even sure how to approach such a proof.

REPLY [13 votes]: Besides the other existing answers, I just want to add that we never in practice need the full axiom of choice. Rather, we typically need just a well-ordering of some relevant objects. For example, we can prove without using AC that every vector space whose vectors can be well-ordered has a basis. Applying that here, we only need a well-ordering of the reals in order to obtain without using AC an additive non-linear function on the reals. What does that mean? Well, to deny the existence of such a function implies denying the existence of a well-ordering of the reals. So if you want to prove that AC is necessary to prove the existence of such a function, you must at least be able to construct a model of ZF in which the reals cannot be well-ordered, under the assumption that ZF is consistent. That should give an idea of why it is unsurprisingly hard in the first place. (It is harder to construct a model of ZF where the reals cannot be well-ordered, than one where the reals are well-ordered.)
Similarly, we have (without AC):

Every partial-order on a well-orderable set in which every chain has an upper bound has a maximal element. (The proof can be easily adapted from here by using the well-ordering to map each chain to a unique strict upper bound.)
Every well-orderable field has an algebraic closure. (This applies to all countable fields, and curiously we do not need AC to that prove the complex numbers are algebraically closed.)

So it is sometimes interesting to note that although many theorems do rely on AC in full generality, when they are applied to specific sets it comes down to merely depending on a well-ordering of one specific set, far from full AC.