The Axiom of Choice (AoC) in set theory famously gives rise to controversial and counterintuitive theorems. (Examples: Banach-Tarski paradox and existence of non-measurable sets.)

I'm aware of some of the theorems that depend on AoC: All vector spaces have a Hamel basis, all fields have an algebraic closure, every set can be well-ordered.

My question is, why is AoC generally accepted/used, rather than rejected/excluded?

The results that depend on AoC that I've encountered don't seem like "necessary truths." If the axioms of set theory couldn't prove that 2 + 2 = 4, I think we'd all agree that they were insufficient at capturing the essence of mathematical truth; but if they failed to prove that vector spaces of arbitrary dimension space must have a Hamel basis... that just doesn't seem to violate my intuition.

On the other hand, the "issues" that arise due to AoC (like non-measurable sets) require adding lots of baggage and extreme care in order to properly tame (at least in measure theory, where in order to assign sizes to sets, one has to define sigma algebras and formulate the entire theory around them). If we rejected AoC, then the Banach-Tarski paradox wouldn't arise and we wouldn't need to worry about sigma algebras because we couldn't construct non-measurable sets in the first place. And we'd only be losing some results that don't seem necessarily or intuitively true? I'm just curious to hear the other side of this argument.

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    Mathematician's today do not care about "necessary truths", they care about technically tidy and applicable theories. And many handy closure results (like Khan-Banach theorem, algebraic field closures, maximal ideals, etc.) depend on the axiom of choice. The baggage like sigma algebras is far outweighed by the disarray without it, e.g. without AC it is possible to partition continuum into more than continuum of pairwise disjoint nonempty sets. And Banach-Tarski paradox is just a curio, that does not affect much in practice.
    – Conifold
    Commented May 20, 2019 at 5:07
  • Actually, you can construct nonmeasurable sets. For example, there are explicitly constructible sets turn out to be non-measurable if you assume the axiom of constructibility (a.k.a. the "V=L" axiom). It turns out that the existence of non-constructible sets is a necessary condition to reject AoC.
    – user6559
    Commented May 22, 2019 at 1:36
  • @Hurkyl I'd argue that the term "constructible" in the sense of Godel is wildly misleading. Commented May 22, 2019 at 4:05
  • "In mathematics, the axiom of choice, or AC, is an axiom of set theory equivalent to the statement that a Cartesian product of a collection of non-empty sets is non-empty." - do you believe the Cartesian product of a collection of non-empty sets can be empty? I don't. That would be utterly absurd. (And because this is philosophy.se and not math.se, I believe utter absurdity is a fairly valid reason not to believe something) Commented Apr 14, 2021 at 11:46

1 Answer 1


There is a lot of writing both in favor and against AC from a philosophical standpoint - e.g. in favor see Penelope Maddy's Believing the axioms.

However, there are also more mundane issues. I think that, whether or not it's ideal, a key point here is usability.

An answer like this may seem dubiously appropriate at philosophy.stackexchange, but I think it's an important part of the picture - and any philosophy of mathematics that doesn't take into account actual mathematical practice is fundamentally incomplete, in my opinion.

First, I think it's easier to recognize an implicit use of tameness assumptions (like "every set is measurable") than it is to recognize an implicit use of choice. (This is especially true because there are tameness assumptions which are secretly applications of choice, like "every vector space has a basis"). What this means is that adding AC as an axiom to ZF makes it substantially easier to recognize a natural-language proof (that is, a proof as written by human mathematicians, as opposed to a fully-formalized proof) as valid (= demonstrating that the principle in question is actually a consequence of our axioms).

That is, there's a distinction between improving proofs and improving results. Even if we accept that choice leads to undesirable consequences moreso than its negation (which I'm not sure I buy), this doesn't address the issue of whether "proving in ZFC" is an easier/more natural task than "proving in ZF" or "proving in ZF + [tameness property]."

For a concrete example of how theories involving "taming" negations of AC can be hard to use, consider the most common AC-alternative: determinacy and its variants. On the one hand determinacy does prove that every set of reals is measurable, has the Baire and perfect set properties, etc., although these proofs are nontrivial. On the other hand, the justification for determinacy is surprisingly fragile: "every game has a winning strategy" misses the fact that the set of moves is fundamentally important - determinacy for games on the countable ordinals is outright inconsistent with ZF!

So in order to actually get tameness out of determinacy we need to spend some work learning how to use determinacy (which is much harder than learning how to use choice, in my opinion); meanwhile, choice is a much more "global" axiom, and the naive justification for choice isn't actually misleading in contrast with that for determinacy.

Note that this is a different question from whether AC is true (whatever that means), but I think we can't ignore pragmatic concerns in studying mathematical practice.

Now coming back to the italicized third paragraph above, an important question at this point (especially for this site) is whether we can extract from this pragmatic concern an actual philosophical argument or observation.

I think in this case we actually can. When we reject choice on the grounds that it implies the existence of "pathological" objects, we're making the implicit assumption that mathematical objects are more fundamental than mathematical proofs, and I think this is unjustified.

Moreover, when we get down to it the objection to choice that you're making is also significantly pragmatic, at least until we've given further justification: e.g. from a Platonist perspective, why should pathological behavior imply nonexistence?

For what it's worth, there is a precise sense in which the issue of AC doesn't affect "concrete" mathematics.

Shoenfield's absoluteness theorem states that any Pi^1_2 statement true in M is true in N, whenever M and N are models of ZF with the same ordinals. This is a bit technical, but the key points are:

  • Basically every concrete mathematical principle is Pi^1_2 (or indeed much simpler).

  • Godel showed that for every model M of ZF there is another model L^M with the same ordinals such that L^M is a model of ZFC.

As a consequence (via the completeness theorem), every Pi^1_2 theorem of ZFC is a theorem of ZF, and so basically every concrete mathematical principle which is ZFC-provable is also ZF-provable. Similarly, the continuum hypothesis doesn't affect Pi^1_2 principles.

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    Nice answer, and I think it is more than appropriate. Focusing on mathematics as practiced is a big trend in the recent philosophy of mathematics, some are even calling it "the practical turn in philosophy of mathematics", see Giardino. Indeed, the prevailing view seems to be that we should ignore whether AC is "true", because that does not mean anything cogent.
    – Conifold
    Commented May 20, 2019 at 23:04

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