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As I'm sure many of you do, I read the XKCD webcomic regularly. The most recent one involves a joke about the Axiom of Choice, which I didn't get.

The Banach-Tarski theorem was actually first developed by King Solomon, but his gruesome attempts to apply it set back set theory for centuries.

I went to Wikipedia to see what the Axiom of Choice is, but as often happens with things like this, the Wikipedia entry is not in plain, simple, understandable language. Can someone give me a nice simple explanation of what this axiom is, and perhaps explain the XKCD joke as well?

Asaf Karagila
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Tyler
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    The punchline is in the alt-text: "The Banach-Tarski theorem was actually first developed by King Solomon, but his gruesome attempts to apply it set back set theory for centuries." – Jyotirmoy Bhattacharya Oct 11 '10 at 05:25
  • Maybe you could look at this question too: http://math.stackexchange.com/questions/198/arent-constructive-math-proofs-more-sound/1547#1547 – Agustí Roig Oct 11 '10 at 08:42
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    If I have an infinite number of pairs of identical socks and can always pick out one from each pair, there is a way to make pumpkins spontaneously reproduce. – Seamus Oct 11 '10 at 10:37
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    Look at: Wagon, Stan (1994). The Banach–Tarski Paradox. Cambridge: Cambridge University Press. ISBN 0-521-45704-1 – Joseph Malkevitch Oct 11 '10 at 13:53
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    The formulation I like is this: the Cartesian product of two infinite sets is always non-empty (including for uncountably infinite sets). – Matt Calhoun Oct 15 '10 at 16:20
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    @Matt: Not precisely, this does not require AC. The correct formulation is: Cartesian product of any family of nonempty sets is nonempty. – sdcvvc Jan 18 '12 at 17:16
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    https://simple.wikipedia.org/wiki/Axiom_of_choice and http://www.explainxkcd.com/wiki/index.php/804:_Pumpkin_Carving ; Both are great sites for when you just don't get it. – aportr Apr 21 '15 at 15:35
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    This [video about Banach-Tarski](https://www.youtube.com/watch?v=s86-Z-CbaHA) is the most accessible explanation I came across so far. – kasperd Sep 27 '15 at 17:08
  • Essentially, as I understand, the axiom allows one to make the choices in 'parallel', so that the time taken or the length of the proof or whatever (used in order to define the function) is the *max* of the times required for the choices as opposed to the *total* (i.e., AC serves as an 'oracle' of sorts, or allows one to act as though the values have been defined just as they're needed or just before they're verified or just in time.). – Vandermonde Sep 21 '16 at 22:18

8 Answers8

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The joke is really about the Banach-Tarski theorem, which says that you can cut up a sphere into a finite number of pieces which when reassembled give you two spheres of the same size as the original sphere. This theorem is extremely counterintuitive since we seem to be doubling volume without adding any material or stretching the material that we have.

The theorem makes use of the Axiom of Choice (AC), which says that if you have a collection of sets then there is a way to select one element from each set. It has been proved that AC cannot be derived from the rest of set theory but must be introduced as an additional axiom. Since AC can be used to derive counterintuitive results such as the Banach-Tarski theorem, some mathematicians are very careful to specify when their arguments depend on AC.

Here is a formal statement of AC. Suppose we have a set $W$ and a rule associating a nonempty set $S_w$ to each $w \in W$. Then AC says that there is a function $$f:W \to \bigcup_{w \in W} S_w$$ such that for all $w \in W$ $$f(w) \in S_w$$

tkp
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Jyotirmoy Bhattacharya
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    one should also try to read about nonmeasurable sets http://en.wikipedia.org/wiki/Nonmeasurable_set – BBischof Oct 11 '10 at 05:23
  • Look at:Wagon, Stan (1994). The Banach–Tarski Paradox. Cambridge: Cambridge University Press. ISBN 0-521-45704-1 – Joseph Malkevitch Oct 11 '10 at 13:51
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    Even though I didn't follow the details, I remembered what the Banach-Tarski theorem was, well enough to understand a visual joke in tonight's season premiere of Futurama, eight months later! Thanks! – Tyler Jun 24 '11 at 05:45
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It might be a good idea to say something about the connection between the axiom of choice and the Banach-Tarski paradox. The reason the Banach-Tarski paradox is counterintuitive is that we expect that if you split up a sphere into finitely many pieces, the total "mass" of all the pieces is still the same - so you shouldn't be able to put those pieces together again into something of twice the total "mass." The reason this reasoning doesn't apply is that the pieces in question don't have mass at all! This is not the same as saying that they have zero mass. It's something much more terrifying: the notion of "mass" (which to a mathematician generally means something precise called measure) can't be defined for these pieces (in a way that still preserves all the reasonable properties we want of our notion of mass).

What does this have to do with the axiom of choice? Well, it turns out that the axiom of choice is one way we can construct these bizarre pieces (non-measurable sets). Without the axiom of choice, it's not possible to prove that such pieces exist. With the axiom of choice, we can construct things like the Vitali set and like the pieces that occur in the Banach-Tarski paradox because AC greatly increases our ability to write down weird sets.

Qiaochu Yuan
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  • Another way of loking at it is to realize that given a set A having volume 1 and a set B having volume 2, both sets have the same cardinality, i.e., both are composed of the same number of points. – David R Tribble Oct 11 '10 at 17:11
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    @Loadmaster: it's more subtle than that. That explains why if you can break A up into infinitely many points and B up into infinitely many points, you can turn one into the other. But the Banach-Tarski paradox accomplishes this with *finitely* many pieces. – Qiaochu Yuan Oct 11 '10 at 17:14
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    @Yaun: Yes, I realize that. I was merely pointing out that even *without* Banach-Tarski, our intuition about infinite sets can be deceiving. Consider a sphere with radius 1 and another with radius 2; both have different volumes but the same cardinalities. Banach-Tarski just takes that same non-intuitiveness to a higher level. – David R Tribble Oct 12 '10 at 22:37
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    Sure you know this, but Banach-Tarski is even **more** terrifying: you can split the sphere into finitely many pieces and assembling them together back forming two spheres of the same size as the original one **isometrically**; that is, whitout any kind of deformation of the pieces. (And, if I remember correctly, you can do that with just four -five?- pieces, one of them being just a point!) – Agustí Roig Oct 16 '10 at 01:19
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    @Agustí Roig: My recollection is that it's five pieces: the center and four non-points. – Charles Oct 21 '10 at 14:46
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    @d.t.: I submit that it's not really that terrifying to anyone who is familiar with real mathematics. After all, you can cut the unit circle into **two** parts and merely rotate one of them such that the two parts are still disjoint but their union now has infinitely many points missing. Hint: {1/2},1+{1/2,1/3},2+{1/2,1/3,1/4},... – user21820 Feb 09 '19 at 16:34
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$\bf 1.$ The Axiom of Choice

Given a set $S$, to say that $S$ is not empty is to say that $\exists x(x\in S)$ (in English: there exists some $x$ such that $x$ is an element of $S$). First-order logic has an inference rule which allows us to move from $\exists x(x\in S)$ to some [new] constant symbol $s$ such that $s\in S$.

This process, called existential instantiation, allows us to move from "the set is not empty" to "here is an element of the set". And this is, in effect, means that we [or rather the inference rule] chose some arbitrary element from $S$.

Suppose now that you have $S_0,S_1$ and neither is empty, then you apply the process twice, and you have two elements of $S_0$ and $S_1$ respectively.1 But what happens if we are given some indexed family $S_n$ for $n\in\Bbb N$ and the information that neither of the $S_n$ is empty?

We cannot use existential instantiation infinitely many times. Remember that mathematics, formally, is always on its way to prove something. Proofs are finite in nature, so you can only apply the inference rule for finitely many of these $S_n$'s.

And to solve this we use the notion of a "choice function". If we had a function $f$ such that $f(n)\in S_n$ for all $n\in\Bbb N$, then we wouldn't need to apply any existential instantiation on the $S_n$'s, since $f(n)$ would already be some fixed element of $S_n$. And the axiom of choice asserts that such $f$ exists, in the broadest way possible, namely if we are given any indexed family of non-empty sets (regardless to the index set), then it has a choice function.

Now we can apply existential instantiation to the set of choice functions, which we have proved to be non-empty using the axiom of choice, and obtain the wanted function.

In simple words, if so, the axiom of choice says that given any family of non-empty sets $S_i$ for $i\in I$, there exists a function such that $f(i)\in S_i$ for all $i\in I$.

But of course, this doesn't quite explain the joke in that last panel. For this we need to talk about...

$\bf 2.$ The Banach-Tarski Paradox

The axiom of choice is extremely useful, and it seems extremely natural as well. If we are given non-empty sets, then there is a way to choose an element from each set. But the consequences of the axiom of choice can be counterintuitive at first.

One of them, called the Banach-Tarski paradox, states that given a ball in $\Bbb R^3$, we can partition it into five parts, move these parts around without stretching or skewing them, and then reconstruct two balls each one exactly the same as the original ball.

This is truly mind boggling, and a lot of people object to the axiom of choice on the ground that this process shouldn't be possible. But those people often mistake mathematical balls to actual physical balls (or vice versa) and a non-constructive mathematical process with what we can do by hand [or robot] in real life.

The XKCD that you link is playing exactly on that. The character in the last panel has cut through the pumpkin several times, and suddenly there were two pumpkins. Just like in the Banach-Tarski paradox. And the "narrator" character points out that they shouldn't have used the axiom of choice to carve out a pumpkin.


Footnotes.

  1. This is not a complete account of the events, and there are more issues to care about. But I find that getting into them can be confusing, and initially it is a good idea to think about the problem as repeating instantiation.
Asaf Karagila
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    It's been on the edge of my mind to write this answer, for quite some time. I'm glad to have finally taken the time to do so. – Asaf Karagila Apr 20 '15 at 14:23
  • Is the axiom of choice provable from induction when the set is countable? – DanielV Apr 20 '15 at 17:09
  • @Daniel: No. It is provable for finite sets using induction. (This is the point I referred to in my footnote, that induction is used to prove it for finite collections, rather than "writing a long sequence of instantiations". The reasons this much is necessary are beyond the scope of "simple terms" though.) – Asaf Karagila Apr 20 '15 at 17:11
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    Why is it forbidden to use existential instantiation infinitely many times? It seems counterintuitive – seldon Jul 23 '16 at 13:26
  • @mattecapu: Because proofs are finite? – Asaf Karagila Jul 23 '16 at 13:30
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    Well what's wrong with "by existential instantiation, every $S_n$ has an element $s_n$"? I mean, a lot of finite proofs are about infinite objects or tasks. Where's the distinction? – seldon Jul 23 '16 at 13:38
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    @mattecapu: You are confusing a lot of things here. (1) The objects might be infinite, but the proofs are finite. We need the proof to be finite, because the logic we use is finite. Yes, we use finite logic to talk about infinite objects. That is why a lot of things about infinite objects cannot be fully decided in first-order logic. (2) Existential instantiation gives us *an object* which satisfies a property (e.g. being an element of a non-empty set). Here you haven't applied EI, you just argued the sets are non-empty. [...] – Asaf Karagila Jul 23 '16 at 13:43
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    [...] (3) What you did is to appeal to the axiom of choice. Otherwise your proof is infinitely long because you need to appeal, *separately* to EI infinitely many times. Not to mention that what will you do when you need to choose from *uncountably* many objects? – Asaf Karagila Jul 23 '16 at 13:44
  • I'm trying to grasp the issue, I'm probably used to reason assuming AC even if I didn't know that until now. Because to me it's perfectly reasonable to be able to choose from sets I know are non-empty. – seldon Jul 23 '16 at 13:47
  • Does this "disalignment" arise from first-order logic formalization? – seldon Jul 23 '16 at 13:50
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    @mattecapu: Sure, because that's what a combination of choice and EI let you do without troubling you. That's the sign of a good foundation: you're not noticing when you're using it. The main point, again, is that you can only appeal to EI *once* each step of the proof. Using the axiom of choice, you can instantiate a choice function, once, and then you immediately get all the choices you wanted to do. By using the choice function to specify your objects. – Asaf Karagila Jul 23 '16 at 13:51
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    I will try to wrap my mind around that... I'm still not understanding the difference between something like induction and using EI at least countably many times. Like, a lot of times you just "declare" you're doing something infinitely many times, like when you use bisection to prove Bolzano-Weierstrass' theorem. Where's the difference? Is something peculiar to first-order logic? – seldon Jul 23 '16 at 14:00
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    @mattecapu: The difference is that induction is a theorem which allows you to do the following: Prove **one** base step; prove **one** step implication; **conclude** that for every natural number something holds. You only prove *two* things, not infinitely many. – Asaf Karagila Jul 23 '16 at 14:01
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    Hi Asaf, thanks for this awesome answer! This is not the first time I read a comment/answer by you about AC, and I always like them, and this one especially helped me a lot. I always asked myself whether moving from "there exists" to "here is one" was trivial or not. Now, regarding your footnote, where can I learn more about these *other issues*? Any good follow-up links? Or should I post a question here on Math.SE? Or can you recommend me a book? Thank you very much. – Pedro A Feb 11 '17 at 03:09
  • @Hamsteriffic: http://math.stackexchange.com/q/412301/622 and http://math.stackexchange.com/q/85153/622 and http://math.stackexchange.com/q/132717/622 make a good start. The issue here is that ZF proves "finite choice", but "finite" as an internal property, rather than meta-theoretic property. This has the benefit of applying to non-standard models, where there are non-standard finite ordinals. Now, all this might be a bit above your head, but you can find in one of these links an answer of mine where I make this very mistake, and there is a discussion below. [...] – Asaf Karagila Feb 11 '17 at 07:55
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    [...] Or, generally, just learn about Goedel's incompleteness theorem, then about models of ZF, and then it will be much easier to understand. You might want to read about "Skolem's paradox" and what I like to call "The reverse Skolem paradox". @Hamsteriffic – Asaf Karagila Feb 11 '17 at 07:57
  • @mattecapu: "For each $n$, you can choose some $x ∈ S_n$" is logically obviously weaker than "You can construct a function $f$ such that for every $n$ it holds that $f(n) ∈ S_n$". It's an invalid quantifier-swap. BW theorem does not require AC, because you can make the choice unique (for the real line, pick the leftmost half that has infinitely many points), and [recursion can be proven in ZF](https://math.stackexchange.com/a/1893865/21820) via the union of approximations. – user21820 Feb 09 '19 at 16:46
  • But it is true that many introductory courses on real analysis use dependent choice without even mentioning it, which is in my opinion very bad practice (it's a quantifier switch...). See [this thread](https://math.stackexchange.com/q/126010/21820) on how countable/dependent choice is often used in the proof that sequential continuity at a point implies continuity at that point. It is necessary as per Asaf's answer, but see Brian's answer for AC-less proof that sequential continuity everywhere implies continuity everywhere! @PedroA: you may also be interested. – user21820 Feb 09 '19 at 16:52
  • @user21820 Thanks! – Pedro A Feb 09 '19 at 16:53
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The annotation to http://www.irregularwebcomic.net/2339.html has an explanation of the Banach-Tarski paradox in easy to read language. It does ignore some of the technical details, but it covers most of the idea of the construction nicely.

Tobias Kildetoft
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  • +1, but this is really a [link-only answer](http://meta.stackexchange.com/a/8259/147650) at the moment. – Mark Hurd May 20 '16 at 12:59
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You have some boxes, at least one. There may be only one box, seven boxes, or infinitely many boxes.

In each and every box, there are some items, at least one. There may be arbitrarily many items.


Question: Can you pick, at least in principle, exactly one item from each box, without necessarily describing the details or rules of how you select them?

The axiom of choice says that the answer to this question is affirmative.


It might seem quite innocuous, but it turns out that the axiom of choice has groundbreaking consequences in virtually all subfields of mathematics. Some of these consequences are fundamental results, others are mind-boggling paradoxes. The long list includes:

  • [Geometry] Banach–Tarski paradox. (The axiom of choice makes it possible to cut an object into a finite number of pieces in such a weird way that you can reassemble two copies of the same object of the same size!)
  • [Topology] Tychonoff’s theorem.
  • [Algebra] Every non-trivial vector space has a Hamel basis.
  • [Order theory] Zorn’s lemma.
  • [Order theory] Well-ordering theorem and transfinite induction.
  • [Set theory] Countable union of countable sets is countable.
  • [Set theory] Every infinite set has a proper subset with the same cardinality.
  • [Measure theory] Existence of sets that are not Lebesgue measurable.
  • [Functional analysis] Baire’s category theorem.
  • [Functional analysis] Hahn–Banach theorem.
  • [Functional analysis] Every Banach space admits linear maps that are nowhere continuous.
  • [Functional analysis] Krein–Milman theorem.
triple_sec
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  • "You have some boxes, at least one. There may be only one box, seven boxes, or infinitely many boxes. In each and every box, there are some items, at least one. There may be arbitrarily many items. Question: Can you pick, at least in principle, exactly one item from each box, without necessarily describing the details or rules of how you select them? The axiom of choice says that the answer to this question is affirmative." – Wd Fusroy Sep 11 '19 at 06:15
  • Thus if we take all the open subset sets on the real line between the integers, i.e. (0,1), (1,2), (2,3) ... it is very easy to choose one unique item from each subset, e.g. all the midpoints, 1/2, 3/2, 5/2 etc.. On the other hand, if the choices must be RANDOM things become much messier since 99.9999999 ... % of all the points between integer values are transcendental numbers, not rational or algebraic numbers. But, alas and alack, the vast majoiry of transcendental numbers are impossible to specify uniquely with any finite string of symbols. So one cannot choose one randomly. – Wd Fusroy Sep 11 '19 at 06:27
  • In fact, the only thing one can do is give a long list of all the digits of the transcendental choice candidate, but wherever one chooses to terminate such a finite list, it remains impossible to specify which of the uncountably many continuations is the unique point you wish to choose. The only alternative is to choose a point that has some "special" finitely-describable status, like some rational or algebraic number ... Most set theorists do not believe there is a real problem here, or that "randomness" is not necessary. But to me it seems very strange to say you can pick a non-random point. – Wd Fusroy Sep 11 '19 at 06:38
  • @WdFusroy What does it mean for one to pick a point randomly? And why is it a problem? I am genuinely curious :) – Saud Molaib Jan 05 '20 at 05:57
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Jyotirmoy writes that:

[To prove the Banach-Tarski theorem] requires the Axiom of Choice.

This statement needs some qualification. A bit of background: the standard collection of set-theoretic axioms is called ZFC, where "C" stands for the axiom of choice. Deleting the axiom of choice yields a smaller collection of axioms called ZF. Here's the key point: when someone says: "So-and-so theorem requires the axiom of choice" what they're usually meaning is "so-and-so theorem does not follow from the ZF axioms alone."

This applies here. Explicitly:

  1. ZF cannot prove the Banach-Tarski theorem
  2. ZFC can prove the Banach-Tarski theorem
  3. There are collections of axioms for set theory intermediate between ZF and ZFC that can prove the Banach-Tarski theorem (without proving the full-blown axiom of choice). The gory details are as follows. In the presence of the ZF axioms:
    • The axiom of choice can be used to prove the ultrafilter lemma, but not conversely.
    • The ultrafilter lemma can be used to prove the Hahn-Banach theorem, but not conversely.
    • The Hahn-Banach theorem can be used to prove the Banach-Tarski theorem.

So, in the presence of the ZF axioms, there's really a range of axioms between the axiom of choice and nothing; somewhere in amongst that tumult is the Banach-Tarski theorem, and it is neither right at the top (read: equivalent to the axiom of choice) nor right at the bottom (read: equivalent to the empty condition.)

goblin GONE
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Here is another equivalent of the axiom of choice which I believe has not yet been discussed. Given a surjective function $f: A \rightarrow B$, there exists a function $g: B \rightarrow A$ such that $f \circ g = Id_{B}$. The idea is the following:

(1) Let $A,B \neq \emptyset$.

(2) Then given that $\forall x \in B \ \exists y \in A \ (f(y) = x)$, by the surjectivity of $f$, we can have a function $g$ such that $g(x) = y$ or $\exists z (g(x) = z \in y)$ and ignore any elements of $A$ that are left-over after choosing these $y$'s.

(3) This allows us to show that $f \circ g$ is an identity function with respect to elements of $B$.

letsmakemuffinstogether
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"[i]f you have a collection of sets then there is a way to select one element from each set." What does "a way" mean? That there is an algorithm that can select such an element? Obviously, there is a non-algorithmic way to do so -- just go to each set and pick out an element from it. That cannot be what AOC means. But if we mean by "function," just a laundry list that associates one thing with another, that is what it comes to.

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    That is, in fact, exactly what AC means; while it's 'obvious' that there's a non-algorithmic means to do it, it's not _true_! You can 'go to each set and pick out an element from it' a _finite_ number of times, but for infinite collections of sets that isn't necessarily so. (And you can't simply say 'pick the smallest element of each set' either; AC is _equivalent_ to the notion that 'every set can be given an order'.) – Steven Stadnicki Dec 09 '12 at 18:05