Why worry about the axiom of choice?
How I Learned to Stop Worrying and Love the Axiom of Choice
The universe can be very a strange place without choice. One consequence of the Axiom of Choice is that when you partition a set into disjoint nonempty parts, then the number of parts does not exceed the number of elements of the set being partitioned. This can fail without the Axiom of Choice. In fact, if all sets of reals are Lebesgue measurable, then it is possible to partition $2^{\omega}$ into more than $2^{\omega}$ many pairwise disjoint nonempty sets!
The best answer I've ever heard --- and I think I heard it here on MathOverflow from Mike Shulman, which suggests that this question is roughly duplicated somewhere else --- is that you should care about constructions "internal" to other categories:
- For many, many applications, one wants "topological" objects: topological vector spaces, topological rings, topological groups, etc. In general, for any algebraic gadget, there's a corresponding topological gadget, by writing the original definition (a la Bourbaki) entirely in terms of sets and functions, and then replacing every set by a topological space and requiring that every function be continuous.
- A closely related example is that you might want "Lie" objects: sets are replaced by smooth manifolds and functions by smooth maps.
- Another closely related example is to work entirely within the "algebraic" category.
In all of these cases, the "axiom of choice" fails. In fact, from the internal-category perspective, the axiom of choice is the following simple statement: every surjection ("epimorphism") splits, i.e. if $f: X\to Y$ is a surjection, then there exists $g: Y \to X$ so that $f\circ g = {\rm id}_Y$. But this is simply false in the topological, Lie, and algebraic categories.
This leads to all sorts of extra rich structure if you do algebra internal to these categories. You have to start thinking about bundles rather than products, there can be "anomalies", etc.
Update:
In the comments, there was a request for a totally explicit example, where Axiom of Choice is commonly used but not necessary. Here's one that I needed recently. Let $\mathcal C$ be an abelian tensor category, by which I mean that it is abelian, has a monoidal structure $\otimes$ that is biadditive on hom-sets, and that has a distinguished natural isomorphism $\text{flip}: X\otimes Y \overset\sim\to Y\otimes X$ which is a "symmetry" in the sense that $\text{flip}^2 = \text{id}$. Then in $\mathcal C$ is makes sense to talk about "Lie algebra objects" and "associative algebra objects", and given an associative algebra $A$ you can define a Lie algebra by "$[x,y] = xy - yx$", where this is short-hand for $[,] = (\cdot) - (\cdot \circ \text{flip})$ — $x,y$ should not be read as elements, but as some sort of generalization. So we can makes sense of the categories of $\text{LIE}_{\mathcal C} = $"Lie algebras in $\mathcal C$" and $\text{ASSOC}_{\mathcal C} = $"associative algebras in $\mathcal C$", and we have a forgetful functor $\text{Forget}: \text{ASSOC}_{\mathcal C} \to \text{LIE}_{\mathcal C}$.
Then one can ask whether $\text{Forget}$ has a left adjoint $U: \text{LIE}_{\mathcal C} \to \text{ASSOC}_{\mathcal C}$. If $\mathcal C$ admits arbitrary countable direct sums, then the answer is yes: the tensor algebra is thence well-defined, and so just form the quotient as you normally would do, being careful to write everything in terms of objects and morphisms rather than elements. In particular, if $\mathfrak g \in \text{LIE}_{\mathcal C}$, then $U\mathfrak g \in \text{ASSOC}_{\mathcal C}$ and it is universal with respect to the property that there is a Lie algebra homomorphism $\mathfrak g \to U\mathfrak g$.
Let's say that $\mathfrak g$ is representable if the map $\mathfrak g \to U\mathfrak g$ is a monomorphism in $\text{LIE}_{\mathcal C}$. By universality, if there is any associative algebra $A$ and a monomorphism $\mathfrak g \to A$, then $\mathfrak g \to U\mathfrak g$ is mono, so this really is the condition that $\mathfrak g$ has some faithful representation. The statement that "Every Lie algebra is representable" is normally known as the Poincare-Birkoff-Witt theorem.
The important point is that the usual proof — the one that Birkoff and Witt gave — requires the Axiom of Choice, because it requires picking a vector-space basis, and so it works only when $\mathcal C$ is the category of $\mathbb K$ vector spaces for $\mathbb K$ a field, or more generally when $\mathcal C$ is the category of $R$-modules for $R$ a commutative ring and $\mathfrak g$ is a free $R$-module, or actually the proof can be made to work for arbitrary Dedekind domains $R$. But in many abelian categories of interest this approach is untenable: not every abelian category is semisimple, and even those that are you often don't have access to bases. So you need other proofs. Provided that $\mathcal C$ is "over $\mathbb Q$" (hom sets are $\mathbb Q$-vector spaces, etc.), a proof that works constructively with no other restrictions on $\mathcal C$ is available in
- Deligne, Pierre; Morgan, John W. Notes on supersymmetry (following Joseph Bernstein). Quantum fields and strings: a course for mathematicians, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. MR1701597.
They give a reference to
- Corwin, L.; Ne'eman, Y.; Sternberg, S. Graded Lie algebras in mathematics and physics (Bose-Fermi symmetry). Rev. Modern Phys. 47 (1975), 573--603. MR0438925.
in which the proof is given when $\mathcal C$ is the category of modules of a (super)commutative ring $R$, with $\otimes = \otimes_R$, and, importantly, $2$ and $3$ are both invertible in $R$. [Edit: I left a comment July 28, 2011, below, but should have included explicitly, that Corwin--Ne'eman--Sternberg require more conditions on $\mathcal C$ than just that $2$ and $3$ are invertible. Certainly as stated "PBW holds when $6$ is invertible" is inconsistent with the examples of Cohn below.]
Finally, with $R$ an arbitrary commutative ring and $\mathcal C$ the category of $R$-modules, if $\mathfrak g$ is torsion-free as a $\mathbb Z$-module then it is representable. This is proved in:
- Cohn, P. M. A remark on the Birkhoff-Witt theorem. J. London Math. Soc. 38 1963 197--203. MR0148717
So it seems that almost all Lie algebras are representable. But notably Cohn gives examples in characteristic $p$ for which PBW fails. His example is as follows. Let $\mathbb K$ be some field of characteristic $p\neq 0$; then in the free associative algebra $\mathbb K\langle x,y\rangle$ on two generators we have $(x+y)^p - x^p - y^p = \Lambda_p(x,y)$ is some non-zero Lie series. Let $R = \mathbb K[\alpha,\beta,\gamma] / (\alpha^p,\beta^p,\gamma^p)$ be a commutative ring, and define $\mathfrak g$ the Lie algebra over $R$ to be generated by $x,y,z$ with the only defining relation being that $\alpha x = \beta y + \gamma z$. Then $\mathfrak g$ is not representable in the category of $R$-modules: $\Lambda_p(\beta y,\gamma z)\neq 0$ in $\mathfrak g$, but $\Lambda_p(\beta y,\gamma z)= 0$ in $U\mathfrak g$.
Yes, many people continue to fuss about the Axiom of Choice.
At least part of the explanation for why people continue to fuss as they do over the Axiom of Choice is surely the historical fact that there was a period of several decades during which the axiom was not known to be relatively consistent with the other axioms of set theory. It was after all not until 1938 that Goedel proved the relatively consistency of ZFC over ZF, using the constructible universe, and several more decades passed until Paul Cohen completed the independence proof by proving that ¬AC is also relatively consistent with ZF, using the method of forcing in 1962. It was during these intermediate times, and especially the time before 1938 when the axiom was not known to be consistent, that the increasingly bizarre consequences of AC were being discovered, and so the habit naturally developed to pay close attention to when the axiom was used. This habit surely lessened after the independence results, but it was not dropped by everyone. And so today mathematics is populated by large numbers of mathematicians like yourself (and perhaps myself), who freely use AC without worry, and who may even find the possibilities occuring in non-AC situations, such as infinite Dedekind finite sets, to be even weirder than the supposed non-regularities of AC, such as the existence of non-measurable sets.
Yet, even though I largely agree with the feeling you indicate in your question, there is still some reason to pay attention to AC. First, in mathematical situations where one can prove the existence of a mathematical structure without AC, then important consequences often follow concerning the complexity of that structure. An explicit construction, even if more complicated that a pure existence proof from AC, often carries with it computational information concerning the nature of the object constructed, such as whethere it is analytic or Borel or $\Delta^1_2$, and so on, and these complexity issues can affect other arguments concerning measurability and whatnot. That is, by nature the non-AC constructions are more explicit and these more explicit argument often carry more information.
But second, there remain certain parts of set-theoretic investigation that only make sense in non-AC contexts. The Axiom of Determinacy, for example, stands in contradiction with the Axiom of Choice but nevertheless contains some fascinating, profound mathematical work, pointing to a kind of mathematical paradise, in which every set of reals is Lebesgue measurable, every set has the property of Baire and the perfect set property. This axiom leads to an alternative vision of what set theory could be like. The possibilities of AD place limitations on what we can expect to prove in ZFC, in part because we expect that there are set-theoretic universe close to our own where AD holds. That is, we are interested in AD even if we believe fundamentally in AC, because we can construct the universe L(R), where AD could hold, even if AC holds in the outer universe V. In order to understand L(R), we need to know which theorems we can rely on there, and so we need to know where we used AC.