Advantage of accepting the axiom of choice

Solution 1:

Well, there is quite the long list:

1. Every set can be well ordered, so every two cardinalities are comparable.
2. Continuity by $\epsilon$-$\delta$ is equivalent to continuity via sequences.
3. There are free ultrafilters.
4. Countable unions of countable sets are countable; similarly cardinal arithmetic become definable for infinite summations and products.
5. Hahn-Banach theorem.
6. Product of compact spaces is compact.
7. Second countable implies Lindelof.
8. Every vector space has a basis.
9. Every unital ring admits a maximal ideal.
10. Every set can be given a group structure.

The list goes on and on and on. In short, however, the main reason is that mathematics has grown to accept the axiom of choice and its consequences.

It really depends on what you want to do in mathematics, if you don't do set theory you're likely to benefit more from the axiom of choice than other axioms. If you are a set theorist (or planning on becoming one) then there will be places where the axiom of choice works naturally for your advantage and others where you will know to drop it and adopt other axioms instead.

Edit: Some of the implications above do not require the entire power of the axiom of choice. Indeed some of them follow by "mere" countable choice (choice for countable families) and other follow from weaker assertions like the Ultrafilter Lemma.

Furthermore, in some cases it is also possible to do things "by hand" and prove existence of ultrafilters, ideals, basis, etc. Most of classical analysis can get away with relatively weak principles like Dependent Choice, however it can require a bit more if you want to talk about $\ell^\infty$ kind of spaces (for example, without the axiom of choice it is possible for $\ell^1$ to be reflexive).

It is also important to notice that the universe can behave as though you have the full axiom of choice for sets which are so big that you will never go beyond their size - but after that thing breaks down badly, so while the axiom of choice is negated in a very strong way the universe that most mathematician like to think about has the axiom of choice in full.

Edit II: Having a few minutes to spare, I thought I should give my two cents why people prefer it (except for the fact they got used to it by now). One of the greatest set theorists of our time once told me during a break from a class he gave that a good axiomatic system is one that you use without noting.

For example, the axiom of extensionality makes a lot of sense because we want two sets to be equal exactly when they have the same elements. So does the axiom of power set.

The axiom of choice is very natural in the sense that it allows us to extend "finite definitions" to infinite ones (a good example is that every vector space has a basis, we can write one down for finite dimensional spaces, but we cannot "write it" for every infinite dimensional space). In this sense we are very accustomed to its gentle grace and willing to accept the few peculiarities like the Banach-Tarski paradox, or a well ordering of the real numbers.

This is why some of the mathematicians who were opposed to the axiom of choice (Lebesgue, Borel) actually used weak versions of it. You just don't notice that it's there.

Further reading on the site:

1. Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?
2. Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?
3. Does every set have a group structure?
4. A question about cardinal arithmetics without the Axiom of Choice
5. axiom of choice: cardinality of general disjoint union
6. Cardinality of an infinite union of finite sets
7. Axiom of choice and automorphisms of vector spaces
8. Axiom of choice - to use or not to use
9. Motivating implications of the axiom of choice?

Solution 2:

One high level (non-mathematical) reason for accepting the Axiom of Choice over, say, the Axiom of Determinacy is that the Axiom of Choice is somehow more "basic." At some level it is more natural to think that all Cartesian products of nonempty sets are nonempty. It doesn't quite seem as natural to think that the game $G_A$ is determined for all $A \subseteq \omega^\omega$.

At least, I think that most mathematicians have come to grips with the idea that there are badly behaved subsets of $\mathbb{R}$: there are sets which are not Lebesgue measurable; there are sets which do not have the Baire property; etc. Why shouldn't there be sets of reals whose associated game is undetermined?

Of course, 100 years ago (or thereabouts) some mathematicians were up in arms over the possibility of well-ordering $\mathbb{R}$, and Zermelo's theorem was considered by some as a great argument against Choice. That is to say, attitudes may change.

Solution 3:

A practical reason is that certain mathematical theorems are known to be equivalent to the axiom of choice. In particular, the fact that every vector space has a basis and the theorem of Tikhonov that arbitrary products of compact spaces are compact require the axiom of choice.

More importantly, th axiom of choice is equivalent to th arbitrary product of nonempty sets being nonempty. This seems to be a consequence of our intuitive notion of set. So even though w seldom use the full axiom of choice, it seems to be part of the way of thining about sets.