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I'm teaching myself axiomatic set theory and I'm having some trouble getting my head around the axiom of choice. I (think I) understand what the axiom says, but I don't get why it is so 'contentious', which probably means I haven't yet digested it properly.

As far as I can make out, one phrasing of the axiom is: for any family of non-empty, pairwise disjoint sets, there exists a set containing exactly one element from each set in the family.

If that's all the axiom states, why is there so much debate around it? If it were stated as there exists a procedure for constructing such a set, that might help me understand (though is that an incorrect statement of the axiom?), but then again:

To use Russell's classic shoes-and-socks example, why won't a coin flip for each pair of socks suffice?

I'm sure this must be a stupid question, but please help me understand why.

Daniel Fischer
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MGA
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    How can you flip a coin uncountably many times? – Umberto P. Mar 17 '14 at 15:41
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    Immediate upvote for the question title – user132181 Mar 17 '14 at 15:42
  • @UmbertoP Isn't the set of pairs of socks countable? – MGA Mar 17 '14 at 15:44
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    @MGA, the axiom of choice applies to any family of sets, no matter the size. Moreover, it's not the statement (which is pretty intuitive as you've noticed) that is contentious, it's the things you can derive as a result and the fact that it does not follow from the other standard axioms of set theory that cause people to question it. – Santiago Canez Mar 17 '14 at 15:44
  • @SantiagoCanez I'm going to naively say "can't you always pick at random?", but I'm suspecting the answer is going to be along the lines of "sure, as long as you assume AC". :-) – MGA Mar 17 '14 at 15:48
  • @MGA, what does "at random" mean? ;) – Santiago Canez Mar 17 '14 at 15:48
  • As said by @Santiago Canez, the axiom "in itself" is not wexed ... For quite a lot time mathematicians used it without notice. Some of its consequences are "implausible" : see Banach-Tarski paradox. About your suggestion, you must take into account that Russell's shoes ans socks example is (of course) very nice, but in math you may find situations where is not easy at all to "imagine" a rule; $AC$ gives you the "reassurance" taht ... also when you do not "see it", there is a choice function waiting for you ... – Mauro ALLEGRANZA Mar 17 '14 at 15:56
  • @MauroAllegranza That makes a lot of sense; however, wouldn't the negation of AC be exactly as "implausible" as the "weird" results that can be derived using it, if not more? – MGA Mar 17 '14 at 16:11
  • I'm not a specialist in set theory; about $AC$, you can read this post. My understanding is that there are a lot of "useful applications" (of course, inside mathematics) of it, that greatly outrun the "unplausible" consequences. – Mauro ALLEGRANZA Mar 17 '14 at 16:22
  • Is it true however that "for practical purposes" (e.g. physics or computer science) you can do away with AC? That would be quite surprising for me. – MGA Mar 17 '14 at 16:29
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    For "practical purposes" (physics included) you can "do away" of a lot of things : rational numbers are enough for measuring processes (with "unlimited" precision); at most, you must use computable real numbers (that are still countable, as the rationals). Of course you can "do away" of set theory as a whole : (@Asaf : I beg your pardon !) I think that set theory is the only math theory devoided of any "applications" outside mathematics... – Mauro ALLEGRANZA Mar 17 '14 at 16:33
  • @MauroAllegranza Very insightful points, thanks. And I learned what a "computable number" is :) – MGA Mar 17 '14 at 16:42
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    I first read this as "why can't you pick stocks using coin flips?" – CoderDennis Mar 17 '14 at 18:07
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    @CoderDennis But of course you can pick stocks that way, and probably better than any Expert Advisor can do for you :-) – Carl Witthoft Mar 17 '14 at 18:37
  • @SantiagoCanez If AC did follow from the other axioms, we wouldn't need it! – David Richerby Mar 18 '14 at 21:45

6 Answers6

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Coin flips don't work because you need to decide which sock goes for "heads" and which one for "tails". Once you've made that assignment you don't need the coin anymore; just assume you always get heads.

vadim123
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  • Ah, because assuming you randomly choose which pair is heads, adding a coin flip is unnecessary, because it would randomly choose a randomly designated object (50% chance to be assigned heads in the first place). – The Ugly Mar 18 '14 at 18:08
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    The problem is that how do you "randomly" choose which sock is heads? It's like Zeno's dichotomy paradox. – vadim123 Mar 18 '14 at 18:29
  • What about if you use a nonrandom method to assign heads and tails? – hippietrail Mar 19 '14 at 02:12
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    @hippietrail: Since random is not well-defined, what is not random? And why would such process exist anyway? If you can make these assignments, random or otherwise, then you can choose from the pairs. But the point is that you can't really do that in the general case without using the axiom of choice. – Asaf Karagila Mar 19 '14 at 02:18
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The sets in the "socks example" are such that you cannot possible make a "reasonable distinction" between the two.

It's more than just that. The union of the pairs, while a countable union of pairs, is not countable. It can be made so that it doesn't even have a countably infinite subset (and sometimes it is possible that there is such countably infinite subset).

On the other hand, the union of the pairs $\{H_n,T_n\}$, where $H_n$ and $T_n$ are the possible outcomes of the $n$-th coin flip, is countable. We can simply map $H_n$ to $2n$ and $T_n$ to $2n+1$. This is an easy injection from the possible outcomes into the natural numbers.

To make this slightly more visual, if I will give you countably many sets of pairs of ants, you will be able to examine each pair and discern its elements, but looking for afar, you will not be able to do that for all the pairs at the same time. Similarly here, in this case, you can examine finitely many sets and discern between each set's elements, but you can't do that uniformly for all the pairs.

Asaf Karagila
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  • Every union of countably many countable sets (and even more for finite ones) is countable. – Paŭlo Ebermann Mar 17 '14 at 19:40
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    If you assume the axiom of choice, sure. If not, then no. – Asaf Karagila Mar 17 '14 at 19:46
  • Really? It seems like I can take the first of the first set (first stripe), 1st of 2nd & 2nd of 1st (2nd stripe), 1st of 3rd, 2nd of 2nd, 3rd of 1st (3rd stripe), etc. – phs Mar 18 '14 at 00:01
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    @phs: But that "etc." is exactly where the axiom of choice gets involved in the process. And if you are dealing with "pairs of socks", then the axiom of choice fails. So the process eventually fails. – Asaf Karagila Mar 18 '14 at 10:23
  • Where are the socks? I'm not disputing the general need for AoC, but in this case (countably many countable sets) there clearly is a bijection from the naturals to the union. Consider enumerating the individual member sets as (countably infinite) rows in a matrix, and then enumerate it like this. The off-diagonals have a natural ordering and finite size, so there is no problem reading them off one at a time. – phs Mar 18 '14 at 18:12
  • @phs: You have to choose for each of the pairs which element is the first, and which is the second. What are these "socks"? I can't fully describe to you, but if you have sufficient knowledge in set theory (in particular forcing), then it is easy to explain how to come up with a model where such "socks" exist, and it's understandable why you cannot uniformly choose which is the first sock and which is the second, for all the pairs at the same time. So no, "quite" obvious is not quite obvious. – Asaf Karagila Mar 18 '14 at 18:16
  • Again, in this example (countably many countable sets), the ordering of the individual sets provides an ordering to choose from. Again, I am not disputing the general need for AoC: I'm just saying it's not needed for this example. – phs Mar 18 '14 at 18:20
  • @phs: What is this example? A countable set of pairs? – Asaf Karagila Mar 18 '14 at 18:22
  • The one @PaŭloEbermann mentions: the countability of the union of a countable family of countable sets. – phs Mar 18 '14 at 18:23
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    @phs: Then you are wrong. You do need some part of the axiom of choice in order to prove that the countable union of countable sets, or even the countable union of finite sets, is countable. And the axiom of choice gets into the game exactly at the stage where you wrote "etc." in your first comment. – Asaf Karagila Mar 18 '14 at 18:26
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    @phs: In order to write the sets in a matrix you need to be able and say which is the first, second, and so on, in each of the sets. If you have that, then yes the union is countable. But the point of the socks example is that you can't have that without assuming the axiom of choice. Because you have to choose for every given set which element of that set is the first, the second, and so on. And indeed there are many examples of models where the axiom of choice fails and there are countable families of countable sets (finite, of bounded size, infinite, what have you) whose union is not. – Asaf Karagila Mar 18 '14 at 18:29
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    @phs: Maybe your confusion comes from the meaning of the term "countable". It gives you the existence of a bijection with the natural numbers (an enumeration of the set), but it does not give you (choose for you) such a bijection. For one set (or finitely many) existential elimination will hand you a concrete enumeration, but not so for infinitely many sets at a times. – Marc van Leeuwen Mar 19 '14 at 13:11
  • @MarcvanLeeuwen That is the missing link, thank you. – phs Mar 19 '14 at 18:18
  • @Marc: That's not fully correct. You sort of slide the case where the model and the meta-theory disagree on the integers (i.e. non-standard integers), in which case this only holds for the meta-theory integers; but it's true internally for everything the model think is finite. It's an even finer point than the place where choice is used in the above proof. I do agree it's a good intuition, but it's not really the reason why this fails. In fact, it's not the reason at all. (This is quite delicate, and I only understood that about a year ago, and I even made that mistake before.) – Asaf Karagila Mar 19 '14 at 18:57
  • @phs: No, it's not quite the missing link. See my previous comment. – Asaf Karagila Mar 19 '14 at 18:58
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Your instinct is basically correct, you can choose socks arbitrarily ("by flipping a coin") without an axiom that lets you. But only because anything you do is necessarily a finite process.

In fact the Axiom of Choice is not needed for finite sets. Various restricted forms of it are theorems of the other axioms: https://mathoverflow.net/questions/32538/finite-axiom-of-choice-how-do-you-prove-it-from-just-zf

AC is controversial when applied to transfinite sets. To over-simplify, you can think of the "controversy" as specifically being related to the fact that it's equivalent to the Well-Ordering Theorem (which my course called the Well-Ordering Principle, but apparently that's ambiguous in other contexts). Nobody ever disputed that finite and countable sets can be well-ordered, it's the rest that are tricky.

There's a joke (that doesn't entirely stand up to analysis, but does reflect the gut instincts of many), that the Axiom of Choice is obviously true, the Well-Ordering Theorem is obviously false, and Zorn's Lemma is obviously incomprehensible. They're all equivalent.

Community
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Steve Jessop
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  • +1 because I was going to put the joke in a comment, but you've obviated that need. Though I'm not sure what you mean by "doesn't entirely stand up to analysis," since the whole point of the joke is that in this case intuition and logic are at odds. – Kyle Strand Mar 18 '14 at 21:23
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    Also, Wiki claims that the originator of this quote is Jerry Bona. Credit where credit is due. – Kyle Strand Mar 18 '14 at 21:24
  • @KyleStrand: I just meant that if you take it literally you end up with someone saying, "well, the AoC looks false to me, so you can't say it's 'obviously' true". I'm not claiming that standing up to analysis is a desirable property of jokes in general, or that this one suffers the lack :-) – Steve Jessop Mar 19 '14 at 11:24
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    And perhaps also credit to Paul Cohen for proving that it's undecidable in ZF anyway, which helps the joke work :-) – Steve Jessop Mar 19 '14 at 11:27
  • To be fair, mathematicians frequently say that things are "obvious" or "clear" when (to the reader) they are not! – Kyle Strand Mar 19 '14 at 16:20
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A point not emphasized in the answers yet is that the axiom of choice is not about what you (or anyone else) can do, like flipping coins) but about the existence of sets. The other ZF axioms ensure the existence of sets defined in various ways, but they do not ensure the existence of random-looking sets. One role of the axiom of choice is to support the intuitive notion that all sets are available, even ones that we can't define.

Andreas Blass
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  • The idea that the axiom of choice "supports the intuitive notion that all sets are available", while great for a superficial intuitive understanding of the axiom, is utterly false when you dig deeper. Gödel's proof that the axiom of choice is consistent with ZF actually starts with a model of ZF and then strips it down, the exact opposite of what you would intuitively think the axiom of choice would require. – Dustan Levenstein Mar 17 '14 at 18:51
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It's contentious because it gives you an access to the uncountable infinity of real numbers that hasn't been 'earned' through some constructive process like taking a limit. This results in certain seeming paradoxes, such Banach-Tarski. In addition--though this is a little more of a personal bias--there is nothing in the natural world which seems to motivate it, i.e. no result that I'm aware of which is of interest to physics or any other science depends on it. Math for any practical purpose is 'complete' without it.

Matt Phillips
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    This is not quite true. There are some nice examples of theorems often used in physics (including quantum mechanics related theorems) which consistently fail without the axiom of choice. – Asaf Karagila Mar 17 '14 at 16:35
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    As a more concrete example, the Hahn-Banach theorem is a powerful and useful theorem of Hilbert spaces which is commonly used in QM and operator theory. – cody Mar 17 '14 at 17:28
  • Please share concrete examples of this, but the Hahn-Banach theorem does not depend on the AoC according to the wikipedia page. – Matt Phillips Mar 17 '14 at 17:32
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    Specifically, the Wikipedia page says that Hahn-Banach is proved by the ultrafilter lemma. Which sounds like the kind of thing I might have known 15 years ago, but my memory isn't good enough to agree or disagree with Wikipedia's claim ;-) – Steve Jessop Mar 17 '14 at 18:38
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    The Hahn-Banach theorem is not provable without some fragment of the axiom of choice. Arzela-Ascoli too. – Asaf Karagila Mar 17 '14 at 18:53
  • Oh, and in any case (again, my source is Wikipedia I'm afraid) Janusz Pawlikowski proved in 1991 that ZF plus the Hahn-Banach theorem is sufficient to prove the Banach-Tarski paradox. So the "actually used by physicists" theorem implies the "physical nonsense" seeming paradox. In order to assert that math for any practical purposes is complete without any of the troubling stuff, you'd have to avoid using Hahn-Banach in physics. Granted, there's no example here of full AoC being needed, but the ultrafilter lemma is as contentious as this answer says AoC is, having the same implication. – Steve Jessop Nov 16 '14 at 13:52
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May I add to Andreas Blass' answer? The axiom of choice is a statement that the set-theoretic universe contains lots of sets. In terms of your example with socks, the question is: If you have a countable number of pairwise disjoint 2-element sets does there exist a set which contains exactly one element from each of these 2-element sets. The axiom of choice says yes, such a set does exists, but there are models of set theory in which the axiom of choice fails and in which no set meets each of the 2-element sets in exactly one element.

user131759
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    Contrariwise you can say that AoC asserts the universe doesn't contain "too many" sets ;-) Goedel's constructible universe satisfies AoC, and doesn't have any sets that aren't part of the set theoretic universe you construct it in, but it excludes any sets that are so complex they don't have choice sets (or can't be well-ordered). – Steve Jessop Nov 16 '14 at 14:02