Published on December 14, 2025 4:08 AM GMT

The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn’s Lemma?

Jerry Bona

I sometimes speak to people who reject the axiom of choice, or who say they would rather only accept weaker versions of the axiom of choice, like the axiom of dependent choice, or most commonly the axiom of countable choice. I think such people should stop being silly, and realize that obviously we need the axiom of choice for modern mathematics, and it’s not that weird anyway! In fact, it’s pretty natural.

So what is the axiom of choice? The axiom of choice simply says that given any indexed collection of (non-empty) sets, you can make one arbitrary choice from each set. It doesn’t matter how many sets you have — you could have an uncountable number of sets. You can still make a choice from all of the sets, saying “I’ll take this element from the first set, that element from the second set, this element from the third set…” and so on.

The axiom of choice is the only explicitly non-constructive axiom of set theory^[1], and for that reason, in the 1920s and 1930s, it was contentious. No longer, however. Almost all modern working mathematicians will accept the axiom of choice without much thought, as they should^[2].

Why do people reject the Axiom of Choice?

When people reject the Axiom of Choice, there’s usually two main examples that they’ll give:

• “But the Banach-Tarski paradox!”
• “But Vitali sets (i.e. non-measurable subsets of the reals)!”

Let’s go through each of these and show you why they’re really not that bad:

Why paradoxes are not that bad

Essentially, what happens with the axiom of choice is that people have heard that there were arguments about non-constructivism in the history of maths^[3] and then learn that the axiom of choice allows a paradox to occur. However, when you really dig into what causes the paradox to occur, it’s to do with the weird nature of other objects, not the axiom of choice. The axiom of choice just gets the blame for unjustified historical reasons. It often allows us to realise that something weird is going on, but it’s not the cause of the weirdness! You’ll see this pattern recur as we go through the examples.

Banach-Tarski Paradox

Banach-Tarski — which I won’t explain in too much detail, for a nice visualisation, see here — is by far the most commonly given reason why people reject the axiom of choice. The Banach-Tarski paradox uses the axiom of choice to prove that if you have a ball (i.e. a non-hollow 3-dimensional sphere), then you can rearrange certain subsets of the ball to create two identical copies of the ball, without adding in any new pieces.

But the axiom of choice is not what causes the weirdness in the Banach-Tarski paradox! This is pretty obvious when you actually read the proof of Banach-Tarski. The axiom of choice simply lets you take the weirdness and construct something out of it geometrically.

The thing that causes something weird to happen — which Banach and Tarski use to build two copies of the unit ball — is that the free group on 2 elements is something called a non-amenable group. That is, there is no consistent way to define a “measure” on the group^[4]. This is where the true weirdness lies, and you can prove the free group on 2 elements is non-amenable within ZF alone. Choice is not necessary for something weird to happen here. It is weird that there are non-amenable groups, but they exist already, without the axiom of choice (in fact, more groups are non-amenable if you reject the axiom of choice than if you accept it^[5]).

Once we have this non-amenable group, we then find a “paradoxical decomposition” of the group in two different ways. See your Honour, the “paradox” has entered the room before choice even got here, so it can’t be guilty! If the group’s not legit, you must acquit!

Banach-Tarski then shows that there’s a copy of this non-amenable group, the free group on 2 elements, within rotations of the 3-dimensional ball^[6], and uses this group to divide the ball into different classes.

Essentially, the classes are defined so that a ~ b iff there is some series of rotations, using this group, that moves a to b. You can then choose one representative from each of these classes. — this is where choice gets used — and then by carefully using the paradoxical decompositions of the free group, two copies of the original ball can be constructed.

All choice is doing here is letting you take the paradoxical behaviour of the free group on 2 elements, and apply it to the sphere. Sure, without choice you would be barred from applying it to the sphere, but that doesn’t really make the paradoxical behaviour go away. You’ve just stopped it from being realised in physical space. Choice is not to blame here! On to count 2!

Vitali sets

Before I can jump into Vitali sets, I’ll first explain the very basics of measure theory (I promise I’ll be quick). A measure is something that basically assigns a “size” to a subset of the real numbers. This “size” should satisfy some nice properties and intuitions that we have about how “size” should work for the real numbers. I’ll quickly define the classic measure, μ, on the real numbers. It works as follows:

1. The measure between any two real numbers, μ([a,b]) is equal to b-a. For example, the measure of [0,1] is 1, the measure of [2,5] is 3. It works exactly the way a fifth grader would tell you it should work.
2. If I shift any set, then the measure stays the same. That is, it doesn’t matter “where” the set is. E.g. if μ(A) = 3, then μ(A + 4) = 3^[7].
3. This measure also has the property that if I ask it, how “big” is the single number {4}, it tells me that the size of that single number is 0. This is true for any single number. Since a single number is just a “point” on a number line, it has no size.

4. This measure also satisfies one other nice property. The key one for our purposes is that if we take two sets A and B with empty intersection^[8], then μ(A∪B) = μ(A) + μ(B).

   In fact, this applies for countable unions of sets, so that if we have countably many sets: A, B, C, …, where no element appears in any two sets, then μ(A∪B∪C∪…) = μ(A) + μ(B) + μ(C) + …

This seems fine, right? Wrong! This is where the Vitali sets come in.

Vitali sets are pathological subsets of the reals. They’re a subset of the reals for which this measure cannot assign a size (it’s not that the size of a Vitali set is 0, there’s no size we can give it at all). To build a Vitali set, we first need to start with the rational numbers. Based on the measure we built above, the set of rational numbers must have size 0. It’s not too hard to see why: the set of rationals is built from countably many individual numbers, and each individual number has measure 0, if you sum up countably many sets of size 0, you end up with a set of… size 0.

This makes sense — we’d naively expect the rational numbers to have size 0. There’s a really tiny amount of rational numbers compared to the reals. It might feel weird because on a number line, you can’t find anywhere a rational number isn’t. But this is the wrong intuition. Everyone agrees the rational numbers should have measure 0.

Now consider the rest of the real numbers. The whole thing. We can start counting all of the real numbers using the rationals to help us! Let’s begin:

1. There are all the real numbers that are rational. Easy!
2. There are all the real numbers that are a rational number plus √2. Great!
3. There are all the real numbers that are a rational number plus π. Also great!

4. There are all the real numbers that a rational number plus Liouville’s constant.

    

   … and so on.

We can break up all the real numbers this way into different “shifts” of the rational numbers. Now, for each “shift,” there are many other shifts that would give us the same collection of real numbers, for example instead of shifting the rationals by π in step 3, I could’ve shifted them by π+1, π+2, π+0.5, etc.

So, let’s make a choice out of all of these shifts that I could’ve written — and that’s where the axiom of choice comes in. For each set, let’s choose a representative between 0 and 1 (so for the π set, we’d choose e.g. π-3). This creates a set which we’ll call the Set Of Crazy Reals.

Now for this Set Of Crazy Reals: the representatives we’ve selected for each of our “shifts;” what is the measure of it? Well, let’s try to use this Set O.C.R. to cover all the reals between 0 and 1.

There’s actually a really easy way to do it! Which is to add all the rational numbers between 0 and 1 to our Set O.C.R. (and subtract 1 if they go over 1). So the numbers between 0 and 1 can be split up by taking:

1. Every element of our Set Of Crazy Reals.
2. Every element of our Set Of Crazy Reals plus 0.5 (-1 if they become too big)
3. Every element of our Set (O.C.R) plus 2/5 (-1 if they become too big)
4. Every element of our Set (O.C.R) plus 123/519 (-1 if…)

and so on. Eventually all the numbers between 0 and 1 will appear in one of these sets, and no two of these sets contain any of the same elements^[9]. Then, if this Set Of Crazy Reals is measurable, we can analyze the results using our measure rule from earlier.

So the set [0,1) is made up of countable unions of our Set Of Crazy Reals, which have just been shifted. If this Set Of Crazy Reals had measure 0, then [0,1) would have to also have measure 0 (since it comes from adding a countable number of sets of measure 0 together). If this Set Of Crazy Reals had measure more than 0, let’s say 0.1; then [0,1) is going to have infinite measure, since we’re adding infinitely many sets^[10] of measure 0.1 together to get all of [0,1). So there’s a contradiction, it can’t be measurable!

Now, where did the weirdness come from here. Well, to me it seems clear that really it came from the fact that the reals can be built out of a bunch of shifted rational numbers, right? But everyone agrees about that. The part where we chose a representative of each of those shifts didn’t seem to add any weirdness, it just realized it. In fact, if we didn’t have the axiom of choice, I think there’d be a weirder consequence. We could still build the reals out of shifted rationals, but there’d be no set to witness how it’s done. If there was no choice, that set would just be banned from existing at all.

You can’t run from your fears forever, choice-haters, you have to overcome them!

Why we should keep the axiom of choice?

Okay, if you’ve read this far, perhaps you’re wondering “Sure, maybe the axiom of choice just lets us realise these monstrosities, but why not banish them? After all, what has the axiom of choice done for me recently?” Let’s go through the list:

• Products of compact spaces are compact (Tychonoff’s Theorem)!
• Every vector space has a basis!
• Every field has an algebraic closure!
• Every surjective map has a right inverse!
• Cardinals are well-ordered!

And:

• The Compactness Theorem in first order logic!!!

If you reject choice, in favour of some sort of abomination like Solovay’s model, then you’ll encounter other, much worse, pathologies. You can’t even state the continuum hypothesis in the traditional sense without the axiom of choice! You can take a product of infinitely many copies of the integers, forming a 2D grid, then a 3D grid, then a 4D grid, until you take a product of infinitely many copies of the integers and then… you can’t say anything about that structure^[11]N. In choice-world, that’s just an infinite-dimensional “grid,” without choice, you can say nothing!

Modern mathematicians all accept the axiom of choice for a reason — it’s extremely useful! It works, and therefore is no longer controversial. Stop blaming the axiom of choice for the pathologies of other sets; what did it ever do to you?

1. Although this is debatable, the axioms of power-set and infinity aren’t that constructive.

2. N.B. on consistency-strength grounds, ZF is consistent iff ZFC is consistent, so there’s no grounds to reject on the grounds of trying to reduce consistency-strength requirements.

3. Or the history of the philosophy of maths — take your pick.

4. Which means, very simply, there is no way to assign to the collection of subsets of G a function μ such that μ(G) = 1 and for any subset of the group A and any element g in the group, μ(A) = μ(gA). For example, for finite groups it is easy to see that they’re amenable, for each subset A, you just take μ(A) = |A|/|G| (the size of A divided by the size of G).

5. Not in the sense that more groups are provably non-amenable, in the sense that fewer groups are provably amenable.

6. Note that you need at least 3 dimensions here, since in 2 dimensions there’s “not enough space” for the free group to be realised.

7. A+4 here is just the set of every element in A plus 4. So if our original set was [0,1] our new set would be [4,5].

8. That is, there is no number that is in A and in B. A and B are entirely “disjoint.”

9. Exercise left to the reader, but you can take it on faith if you prefer!

10. Note for pedants: disjoint sets.

11. ^{^}

    Note, this is not quite right. It only occurs in Solovay’s model for countable infinities, but does occur for uncountable infinities and for certain sets (not the integers, since they have an ordering which allows us to make a choice, the linked post is incorrect about that claim).