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Would someone please explain:

  • What does the Axiom of Choice mean, intuitively?
  • What does the Axiom of Determinancy mean, intuitively, and how does it contradict the Axiom of Choice?

as simple words as possible?

From what I've gathered from the Wikipedia page, my understanding is that the Axiom of Choice pretty much lets you make a completely random decision, i.e. equal probability for everything. I don't, however, understand what they're talking about regarding the sets of sets (is it more than what I just described?), and I further don't understand the Axiom of Determinancy.

Info on those would be appreciated. :)

Asaf Karagila
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user541686
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    The Axiom of Choice is this: If you are in an old fashioned sweetie shop, with some infinitely big sweetie jars, then you can take one sweetie from each jar and put them all in another jar. Just take the top sweetie, you say? Well, that requires that the sweeties in each jar are well-ordered... – user1729 Jul 20 '11 at 07:39
  • @Swlabr: So isn't it pretty much making a random selection? – user541686 Jul 20 '11 at 07:41
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    @Mehrdad: The selection is not random, but *arbitrary*, and even then you can impose some limitations. – Asaf Karagila Jul 20 '11 at 07:46
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    @Asaf: I'm confused... when there's no distinction whatsoever between the elements, then don't "random" and "arbitrary" mean the same thing? (Sorry, I'm not great with the lingo...) How can it be arbitrary but not random (or vice versa)? – user541686 Jul 20 '11 at 07:51
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    @Mehrdad: While it is possible to explain the axiom of choice and the axiom of determinacy in *relatively* simple terms, I would like to know what is your mathematical background and motivation for asking this question? – Asaf Karagila Jul 20 '11 at 07:53
  • @Asaf: As a college undergrad, in linear algebra, my instructor often went off-topic and talked about the Axiom of Choice, making me curious/confused. :-) I do have some background in set theory and whatnot, but as I'm not majoring in math, I don't have a great enough background in really abstract ideas, so I'm trying to get a more intuitive feel for things. – user541686 Jul 20 '11 at 07:54
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    @Mehrdad: Linguistically you are correct. However random has a meaning in the mathematical jargon. Furthermore, sometimes you can in fact specify some characteristics of the elements you want to be picked making it less "random" in comparison to the other elements. The point in the axiom of choice is that you usually don't care about which element it is, just that it exists and that it has a certain property. – Asaf Karagila Jul 20 '11 at 07:55
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    @Asaf: But as soon as you specify a characteristic, then you know what you're looking for, and hence there's no problem picking it, right? Isn't the problem when they all look the same (and so *any* choice would have to be random)? (As a side note, it's making me think of quantum mechanics and [Buridan's donkey](http://en.wikipedia.org/wiki/Buridan's_ass).) – user541686 Jul 20 '11 at 07:57
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    The Axiom of Choice is equivalent to the Well-Ordering Principle. This basically says (I think...) that your choice isn't random. Your elements have some order, so you can take the "biggest" or the "smallest" element. So you can take the top element in your sweetie jar... – user1729 Jul 20 '11 at 08:05
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    @Mehrdad: If you haven't already, it might be worth familiarizing yourself with the rest of the ZF axioms. Looking at AC (and AD) out of context is a bit misleading, since you likely have an intuitive notion of "set" that is a bit more generous than the sets actually granted by the axioms (and you'd be in excellent company if that's the case). – user83827 Jul 20 '11 at 08:11
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    @Mehrdad: When you say you have background in set theory, do you mean *axiomatic* set theory (e.g. ZF, NBG, NF, ETCS) or just simple naïve set theory? There's a world of difference! – Zhen Lin Jul 20 '11 at 09:12
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    possible duplicate of [Can you explain the "Axiom of choice" in simple terms?](http://math.stackexchange.com/questions/6489/can-you-explain-the-axiom-of-choice-in-simple-terms) – Pete L. Clark Jul 20 '11 at 09:23
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    @Pete: I agree there is some relation, however the axiom of determinacy was not address in previous questions; nor was the connection of the two. – Asaf Karagila Jul 20 '11 at 09:32
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    As a sider remark: in an instructor in undergraduate linear algebra is going off tangents talking about axiom of choice, either he is doing something awfully wrong or your linear algebra course is not only very advanced, but probably a course in functional analysis in disguise... – Willie Wong Jul 20 '11 at 13:09
  • @Willie: What will you say about the course if the teacher is talking about AD, then? :-) – Asaf Karagila Jul 20 '11 at 13:51
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    I agree with Willie: part of being a good instructor is showing some restraint. Or, as Lucifer Morningstar says in the first issue of his Vertigo comic book series, *You'd think part of omniscience would be knowing when to stop.* – Pete L. Clark Jul 20 '11 at 14:00
  • @Willie: Well he was actually a TA, and it wasn't THAT bad (e.g. he'd say something like, In order to prove that there's a basis for every vector space, you need the AoC... and start talking about that).. it wasn't outright irrelevant, but still tangential. And yeah, our course was rather introductory, mainly to teach us about vector spaces, bases, and get us ready for linear ODEs. – user541686 Jul 20 '11 at 14:35
  • @Zhen: No, I mean the naive one. :P As much as an engineer would need to know. :) – user541686 Jul 20 '11 at 14:35
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    @Mehrdad: I have not yet seen an engineer that even had to study *any* set theory at all (and no, I do not count learning what are intersections, unions and equivalence relations as naive set theory). – Asaf Karagila Jul 20 '11 at 14:50
  • @Asaf: I think I just explicitly said I meant the naive ones... – user541686 Jul 20 '11 at 14:56
  • @Mehrdad: You also said "As much as an engineer would need to know"... :-) – Asaf Karagila Jul 20 '11 at 15:00
  • @Asaf: So you think no engineer needs to know about the "naive" set theory? (Which isn't true, btw. It helps a lot when you're dealing with Dirac deltas, matrices, and Fourier transforms in EE, for example.) Otherwise, I'm really not understanding what your point is... – user541686 Jul 20 '11 at 15:06
  • @Mehrdad: Set theory, as whole, helps a lot to understand many things in mathematics (in fact, any sort of mathematical skill is helpful to the others). – Asaf Karagila Jul 20 '11 at 15:18
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    Bertrand Russel explained the Axiom of Choice as "Given infinity pairs of socks, you need the axiom of choice to choose one from each pair. Given infinity shoes, you do not, since you can specify 'pick the left shoe each time'." – tomcuchta Jul 20 '11 at 16:00
  • @tomcuchta: Right. My question is, how is that different from, "Given an infinite number of socks (i.e. a single set of socks, not a set of pairs), you need the axiom of choice to choose a single one."? i.e. why do you need pairs? – user541686 Jul 20 '11 at 16:02
  • @Mehrdad: Had you asked that originally my answer would have been shorter. If you have singletons then there exists a unique element in each set you choose from. You do not need the axiom of choice for that. However if you are choosing from pairs of socks then it is impossible to say in a finite sentence how to choose from *each* pairs *exactly* one sock. – Asaf Karagila Jul 20 '11 at 16:17
  • @Asaf: "impossible to say in a finite sentence" -> are we talking about computability? Otherwise I don't understand how that's relevant here. – user541686 Jul 20 '11 at 16:18
  • @Mehrdad: If something can be defined by a sentence (which is finite) from your axioms then it exists in every model simply because its existence if provable. If you cannot formulate a finite sentence which describes the choice function then its existence might not be provable from the axioms. And sure enough it is not provable without the axiom of choice. If you have countably many pairs you can say "Choose this from the first pair, that from the second ..." but you cannot say how to choose from all the pairs at once. The axiom of choice *guarantees* that such way exists in the universe. – Asaf Karagila Jul 20 '11 at 16:23
  • @Asaf: Hm... what I found slightly confusing is, how can you choose "this" one from a pair, when there's no way to distinguish "this" one from "that" one? – user541686 Jul 20 '11 at 16:25
  • @Mehrdad: Suppose $\{a_n,b_n\}$ is the $n$-th pair. In a local fashion you can point out and say "This element is $a_n$ and the other is $b_n$". So you can "I want $a_1,a_2,b_3,a_5,b_8$ to be chosen." Because when given a finite collection you can *always* distinguish between its elements. However if you "zoom out" and consider the infinitely many pairs, then what? Then you cannot distinguish *globally* between $a$'s and $b$'s. This is the entire point in Russell's analogy to socks. – Asaf Karagila Jul 20 '11 at 16:29
  • @Asaf: "Because when given a finite collection you can always distinguish between its elements." -> How/why? If there is no difference between two socks, how can you choose and pick one up in your hand? You can't number them because you can't choose one in the first place (which one do you pick first? they're all the same...) so it's a chicken/egg problem, isn't it? Also, would you mind removing the last comment on your post? People read it because it was upvoted, not realizing that it's completely misinterpreting what I was saying. :\ – user541686 Jul 20 '11 at 16:33
  • @Mehrdad: This analogy *can* be taken into the real world. Assuming you have only a finite number of socks, lay them in pairs on your bed. You are able to say "Aha, this is the first sock of the pair, and this is the second sock of the pair!". If, however, you consider infinitely many socks you cannot possible say which sock is the first and which is the second of each pair *for all the pairs*. There is no "egg/chicken" problem, you are simply applying the wrong kind of intuition on the whole definition of "able to distinguish". I will add more to my answer, in hope that it helps. – Asaf Karagila Jul 20 '11 at 16:37
  • @Asaf: "you are simply applying the wrong kind of intuition on the whole definition of 'able to distinguish'" --> it's probably very well the case, since the first thing that comes to my mind is quantum mechanics (specifically, the book The Quantum Zoo, which has explanations on how being able to distinguish two things actually affects their probability) and so my intuition doesn't apply to math here. :-) Thanks for the response! – user541686 Jul 20 '11 at 16:45
  • @Mehrdad: I have added more to my answer. From all the fuss about the axiom of choice I did not hear a word about determinacy and whether or not the axiom was understood. – Asaf Karagila Jul 20 '11 at 17:32
  • @Asaf: Yeah I'm not sure I still get how they contradict, but +1 for the info on your post, at least it explains what it is. :-) – user541686 Jul 20 '11 at 17:44
  • @Mehrdad: Determinacy is a lot more technical than choice, as I say in my answer. I will try to add on this part as well. – Asaf Karagila Jul 20 '11 at 17:48

4 Answers4

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The axiom of choice is an axiom of set theory, and it is used when implicitly assuming that the universe consists of sets, and sets alone. That is - everything is a set.

For example, $0=\emptyset$ and $1=\{\emptyset\}$. Since everything is a set, consider a set $X$ and all its elements are non-empty sets themselves. Formally $\forall y(y\in X\rightarrow \exists z(z\in y))$.

The axiom of choice asserts the existence of a function whose domain is $X$ and its range is $\{z\mid\exists y(y\in X\land z\in y)\}$ (often denoted $\bigcup X$). The function has a very special property, namely $f(y)\in y$. It chooses someone from every element of $X$.

Some examples (which do not require the axiom of choice) are, $X=\mathcal P(\mathbb N)\setminus\{\emptyset\}$ that is all the non-empty subsets of natural numbers.

We can have $f(A)=\min A$, that is the least element of $A$ is returned. We can require something slightly more peculiar $f(A) = \min\{a\in A\mid a\text{ is even}\}$ if such $a$ exists, or $\min A$ otherwise.

These are two examples which do not require the axiom of choice since we are able to specify in a uniform formula who do we want to pick from every $A$.

A useful equivalence of the axiom of choice is Zorn's lemma. This lemma is somewhat complicated, but it asserts that if $(A,<)$ is a partially ordered set, and every linearly ordered subset of $A$ has an upper bound, then there exists a maximal element.

To elaborate on that, if $A$ is a non-empty set, and $<$ defines a partial order on $A$, if every $C\subseteq A$ which has the property $\forall a\forall b(a<b\lor b<a\lor a=b)$ has some $x\in A$ such that $\forall a(a\in C\rightarrow a<x)$ then there exists a maximal element - some $x\in A$ for which the property $\forall a(a\neq x\rightarrow a<x)$ is true.

Zorn's lemma is very useful in algebra, and used in the proof that every field has an algebraic closure; every vector space has a basis; and every ideal can be extended to a maximal ideal (and many many many other uses).

The proof uses heavily the axiom of choice (not surprisingly, since the two are equivalent) and in a nutshell we take one element, then $\{a_0\}$ is linearly ordered, therefore if $a_0$ is not maximal in $A$ we can extend it. That is the set $\{b\in A\mid a<b\land a\neq b\}$ is non-empty, and we can choose from it. We choose from non-empty subsets of $A$ until we either "find" a maximal element, or derive contradiction to one property or another.

I use "find" because many times the maximal element is one we cannot describe nicely by a sentence (and in fact without the axiom of choice there are vector spaces without basis, fields without algebraic closures and so on).


To understand the axiom of determinacy first we need to understand what there is to be determined.

Take $A\subseteq\mathbb N^\mathbb N$, that is a set of infinite sequences of natural numbers.

Now we play a game, I will be Player I (P-I) and you will be Player I (P-II). I will choose some $n\in\mathbb N$, and then you will choose another. The game is infinitely long and will have another round for every finite number of rounds.

Note that in the $n$-th round we have $x_n$ and $y_n$ (I chose $x$'s and you choose $y$'s). This defines a sequence: $$a_n =\begin{cases}x_k & n=2k\\ y_k & n=2k+1\end{cases}$$

We say that I win the game if $\langle a_0,a_1\ldots\rangle\in A$, otherwise you win the game. If there is a strategy assuring victory for either one of us then we say that the game is determined. This is to say there is a function from $\mathbb N$ to $\mathbb N$ that given the current state of the game will give me a possible tail segment which assures one of the player's victory. (Note that there are usually a lot of possible strategies).

For example $A$ will be all the sequences which are constants $a_n=k$ for some $k$. Clearly you have a winning strategy. Whatever I chose at first, choose something else and I have no chance of winning.

Another example is $A$ will be the set of sequences that $a_n$ is even whenever $n$ is even ($a_n=n$, for example). All I have to do is choose even numbers in my turn, and I cannot lose.

The Axiom of determinacy asserts: Every game is determined. That is, if we play this sort of game then regardless to which set of sequences we have chosen, one of us can win.

The conflict with the axiom of choice is a bit technical for this post (and the part about determinacy could surely be phrased better by someone else), the idea is as such. If we assume the axiom of determinacy, then every game is determined. We will define $A$ by a transfinite induction. Since at each step during the game the collection of winning strategies is non-empty. We simply choose such winning strategy and ensure that it will not work by adding a sequence to $A$, repeating the process "enough" times we ensured that no strategy exists to determine the winner after any finite number of turns. This contradicts the assumption that every game is determined.

Determinacy is somewhat of a technical axiom, and while it was very natural to postulate this axiom in some parts of set theory (namely descriptive set theory), the mainstream mathematics has been made very comfortable with the axiom of choice, and as Theo points out in the comments some pathologies which we are used to are gone when assuming the axiom of determinacy.

This makes the axiom of choice more common in modern mathematics than determinacy. Things can change, though... things can change.


Added:

To understand the need for the axiom of choice, we return to Bertrand Russell's wonderful analogy. Given infinitely many pairs of shoes, you can always choose one from each pair, but to choose a sock from infinitely many pairs of socks you need the axiom of choice.

What does that mean? Well, given shoes you can easily say "Pick all the left shoes", and regardless of how many pairs are given in each pair there is exactly one left shoe. This defines a function which chooses an element of each pair. On the other hand socks are indistinguishable and you cannot say which one is the left and which one is the right.

What does that mean indistinguishable? Well, given one pair of socks $\{a,b\}$ we can always say "Oh, this is $a$ and this is $b$", even if our assignment of $a$ was arbitrary. Every time we are given three pairs of socks we can ad-hoc assign each pair as $a_i, b_i$ and pick the ones we assigned as $a_i$. However if there are infinitely many pairs of socks, can we always make such assignment? Well, only if we assume the axiom of choice holds.

The point is that a pair of socks has two elements in it. These are always distinct and we can always examine the pair by itself and distinguish between the two socks. We can always distinguish between three, four and five socks at once; as well ten or fifteen pairs of socks. We simply assign each collection of socks with different names to be able and temporarily tell which sock is which, then we can choose the names we would like.

Mathematically speaking if we are given a finite collection of non-empty sets, if we do not actually care about the choice of elements, as long as we choose exactly one from each set, it is always doable. First we need to understand that "not caring" means that we only require $f(A)\in A$ which is of course doable since $A$ is non-empty. However, in mathematics it is not enough to claim things, one has to prove them as well. In this case, we can simply write the following statment (assuming $A_1,\ldots, A_n$ are our non-empty sets):

$$\exists x_1\ldots\exists x_n(x_1\in A_1\land x_2\in A_2\land\ldots\land x_n\in A_n\bigwedge f(A_1)=x_1\land f(A_2)=x_2\land\ldots\land f(A_n)=x_n)$$

That is to describe exactly (I am somewhat cheating here, I did not describe it exactly, but I gave a close enough approximation) how $f$ looks like, it is the function (however a function may be defined set theoretically) that after fixing $x_i\in A_i$ simply returns $x_i$ as the choice from $A_i$. Since this is a finite collection of pairs we can describe this sort of choice.

If we have one pair then this sentence is very short, as we keep on adding pairs we will write longer and longer sentences. If we finally have infinitely many pairs then we cannot write this sort of sentence. We need to find a different formulation. This is where the ability to uniformly distinguish some unique element in each $A_i$ comes to help. Suppose $\varphi(x)$ is the property of being a left shoe. Let $B=\{B_n\mid n\in\mathbb N\}$ be a collection of infinitely many pairs of shoes. If we want to choose from each pair, we can simply do it as such:

$$\forall X(X\in B\rightarrow \varphi(f(X)))$$

Since there exists exactly one element, $a$ in every $X$ for which $\varphi(a)$ is true (i.e. there is exactly one left shoe in each pair of shoes) this implies the function $f$ is nicely defined, in a finite (and even short) sentence.

Now we return to the socks. There is no property $\varphi(x)$ such that in every pair of socks there is exactly one sock for which $\varphi$ is true. This means that we cannot write a very nice sentence as above, to choose one sock from each pair!

This is the (with a capital "the") key of the issue here. In a finite collection you can always distinguish every element from every other element. When looking at an infinite collection you may not be able to have this luxury.

What does that mean mathematically? It means that you may not be able to write a finite sentence to help you choose from infinitely many pairs without the axiom of choice - which asserts this sort of choice exists (it may not be computable or even definable except for knowing it exists somewhere in your universe, and therefore you can take an arbitrary one and use it for a while).

Asaf Karagila
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  • Re: last sentence. Isn't this simply due to the fact that adherents of "mainstream mathematics" are trained to a context in which the full-fledged axiom of choice is accepted without much ado and that, while AD has very nice consequences, they are completely orthogonal to the pathologies we're used to? Wouldn't it mean to sacrifice the bulk of 20th century's mathematics to adopt an axiom implying measurability of all sets of reals or continuity of linear operators between Banach spaces? Also, accepting the fact that ZF + AD implies consistency of ZF would take some getting used to, I guess. – t.b. Jul 20 '11 at 09:17
  • By "sacrifice" I mean of course "unaccept them" and "re-prove them" if possible. – t.b. Jul 20 '11 at 09:21
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    This answer is at the wrong level and self-indulgent. And honestly probably doesn't help the original poster. – Apprentice Queue Jul 20 '11 at 11:12
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    @Apprentice: Can you elaborate on this opinion of yours? – Asaf Karagila Jul 20 '11 at 11:15
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    I partly agree with Apprentice Queue. But if the OP doesn't undestand the concept of sets-of-sets, he should consult a basic set theory textbook first (or actually the first chapter of most introductory textbooks to modern abstract algebra), after which Asaf's answer would be at the right level. – Willie Wong Jul 20 '11 at 13:03
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    @Willie: This first chapter can be the first chapter of most modern introductory mathematics books (well, except maybe set theory introductory books... :-)). – Asaf Karagila Jul 20 '11 at 13:06
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    @Willie: I completely understand the concept of sets-of-sets, but I also completely agree with @Apprentice. There was *nothing* **intuitive** about this post -- it lost me the moment it said `1 = {null}` (a number is equal to a set? huh?). So yeah, this was pretty unhelpful, unfortunately, though I appreciate the attempt. – user541686 Jul 20 '11 at 14:39
  • @Willie: If it's confusing what I mean by "intuitive" (how can something technical be intuitive?? It's impossible!!) it shouldn't be. I could likely give you a good intuitive definition of another area of math (Fourier transforms maybe? or how about Fourier series in linear algebra?) or area of science (Heisenberg Uncertainty Principle, anyone?), and it'd hopefully make sense how something complicated could probably be explained in an intuitive way. In general, if you find that you *have* to use something technical to explain something technical, you're probably just... wrong, I'm sorry. :-) – user541686 Jul 20 '11 at 14:45
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    @Mehrdad: I have to say that in order to understand a concept which is purely set theoretic, you need to accept *some* set theoretic ideas. I gave the example that $1$ is *defined* as $\{\emptyset\}$ in order to explain *how* normal mathematics is defined *within* set theory. As my first line says: the axiom of choice is **an axiom of set theory**. – Asaf Karagila Jul 20 '11 at 14:46
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    @Mehrdad: I strongly suggest you to read the meta.MSE thread in http://meta.math.stackexchange.com/questions/2606 and especially Carl's answer there. Intuition can be acquired by studying, if you study the basics of axiomatic set theory, eventually what I wrote becomes intuitive. Without studying there is no intuition. – Asaf Karagila Jul 20 '11 at 14:48
  • @Asaf: And yet, Wikipedia's explanation (with the shoes/socks) didn't have a single piece of math in it (as neither did Swlabr's example), and it at made *some* sense, though I'm asking for clarifications here (e.g. regarding my interpretation regarding randomness, etc.). So I don't believe you "need" something already technical to explain something technical in simple terms, although you certainly *will* need them if you actually want to *apply* the ideas. – user541686 Jul 20 '11 at 14:49
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    @Mehrdad: To be fair I found *nothing* intuitive in Swlabr's example except for the fact he did not use "sets" or "numbers". If you want, I could rewrite my answer and replace "set" by "Kreplach" and element by "meat filling". – Asaf Karagila Jul 20 '11 at 14:59
  • @Asaf: I found it intuitive because it explained that (from what I understood) the AC *allows you to make a single choice out of a bunch of equivalent choices, i.e. when you logically can't distinguish any choice from any other*. I guess before going farther, I should ask: Is this a correct interpretation in the first place? – user541686 Jul 20 '11 at 15:03
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    @Mehrdad: What does it even mean "when you logically can't distinguish..."? The interpretation of the axiom of choice is single: If you have a collection of non-empty sets then you can choose exactly one element from each set. If you have *one* choice function then you have *all* possible choice functions. The use of "equivalent" and "single" is unclear. – Asaf Karagila Jul 20 '11 at 15:06
  • @Asaf: Let me rephrase that: *the AC allows you to make a single choice out of a bunch of choices which have no distinguishing property*. Is it clearer now? – user541686 Jul 20 '11 at 15:09
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    @Mehrdad: Not really, actually. The Axiom of Choice allows you make a choice *at all*, that is the only point. There is (an even more technical) example of a *countable* set of *pairs* which have no choice function. That is there is *no choice whatsoever*, let alone a *single* choice. – Asaf Karagila Jul 20 '11 at 15:15
  • @Asaf: OK. Is this (finally) clear now? *The AC allows you to make **any choice at all** from a bunch of choices which have no distinguishing property.* – user541686 Jul 20 '11 at 15:20
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    @Mehrdad: The point is that you don't *really* care about which choice, but rather that you can make it. Furthermore, within a certain set there is always a distinction between elements. However if you do not have a uniform way to distinguish *one* element from each set, there might not be a way to choose from infinitely many sets at once. – Asaf Karagila Jul 20 '11 at 15:25
  • @Asaf: OK, in that case, then we just don't understand each other, probably because of my lack of expertise. tl;dr: I didn't find this answer helpful. Sorry, and thanks for the input. – user541686 Jul 20 '11 at 15:43
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    @Mehrdad: it cannot be made intuitive. In the realm where your everyday experience (if you are an empiricist and believe that intuition comes from your past interactions with the world) applies, axiom of choice is **never needed**, as the totality of your experience is necessarily finite. And with the types of infinites one deals with in axiomatic set theory, [certain things that shouldn't be true based on "finite" intuition can be made to happen](http://en.wikipedia.org/wiki/Banach-Tarski_paradox). – Willie Wong Jul 20 '11 at 16:49
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    @Mehrdad: if you are interested in a relaxed, layman's account of why axiomatic set theory came to be, and why in dealing with issues of *foundations of mathematics* one often necessarily disgard "intuition", I can recommend the book [LogiComix](http://www.logicomix.com/en/). – Willie Wong Jul 20 '11 at 16:50
  • @Willie: Yeah I've actually heard of that paradox before (also from my TA, mind you, though I'd heard about it earlier as well)... it's beyond what my brain can handle, though, so I didn't ask about it. :-) – user541686 Jul 20 '11 at 16:56
  • Could someone explain @ApprenticeQueue about the answer being wrong? I know just very basic set theory stuff, but most of Asaf's answer made sense. – YoTengoUnLCD Mar 24 '16 at 17:41
  • @YoTengoUnLCD: That happened nearly five years ago. Leave it be. – Asaf Karagila Mar 24 '16 at 18:03
  • @AsafKaragila Whoops... I did not check the date. It wasn't meant to be an attack against you or anything: I just couldn't find any inconsistencies between what you wrote and the little I know, so I was curious. – YoTengoUnLCD Mar 24 '16 at 18:06
  • @YoTengoUnLCD: I don't think this is something against me. But it is a comment from five years ago. – Asaf Karagila Mar 24 '16 at 18:08
  • @AsafKaragila Really like your idea of needing the axiom of choice to *go beyond* the finite sentence! – Atom Apr 23 '20 at 20:31
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I have nothing technical to add to Asaf's answer (he's the one who works on the axiom of choice, after all :-), but something from the perspective of someone who wondered for a long time what the axiom of choice is all about and realized in the end that this confusion arose from a deep misunderstanding of the point of axiomatic set theory. If you already have some (philosophical or pre-philosophical) notion of what it means for mathematical objects like sets to "exist", it happens easily that you read statements about the "existence" of a set (which abound in axiomatic set theory, including Asaf's answer) in the light of that pre-existing notion. That naturally tends to lead to reactions such as: "Whaddaya mean, there might not be a choice function? Here's the set of sets, there's the set of their elements, you agree that both of these exist and that their Cartesian product exists, how can you say there might not be a subset of that product containing exactly one pair for each set relating it to one of its elements? It's right there, that's what the Cartesian product is all about!"

The point is that axiomatic set theory intentionally forgets about all such intuitions of what sorts of sets "there are" and deals only with formal statements about the existence of sets that can be deduced from the axioms and/or that hold in certain models (which often bear little resemblance to what you intuitively think of as "the sets that there are or should be"). If you read that "a choice function might not exist", don't interpret this as telling you something you can or can't do or imagine with infinite sets; it simply means that the existence of a choice function cannot be proved from the axioms, or that there is no choice function in the particular model being discussed.

If you stick to this advice, the axiom of choice becomes a lot less mysterious than it may seem. It's actually quite easy to imagine that there are models of certain axioms that don't contain sets that one thinks "should exist" -- after all, they're just models and axioms. If you don't use the axiom of infinity, you can't deduce from the remaining axioms that there's an infinite set, and you can find a model that doesn't contain any infinite sets -- that has nothing to do with whether infinite sets "really exist" in a philosophical sense of the word (if there is one :-).

Related to this is that it's important to clearly distinguish "internal" and "external" statements -- axiomatic set theory is always concerned with a certain universe of sets, and we can always talk about this "from the outside" and talk about sets that "don't exist" inside the universe under consideration. This would obviously not be possible if all statements about "existence" referred to "the actual universe of sets" (whatever that might be) and not to some particular one.

After all this, one might wonder why bother with axiomatic set theory if it's just a formal game with axioms and models. The reason for this is that our intuition about sets can prove quite unreliable, as evidenced by the paradoxes that pop up everywhere in what's now called "naive set theory". We can say something like "the set of all sets that don't contain themselves", and think that we've said something meaningful, whereas in fact we've said something utterly meaningless, since there can be no such thing. Now that doesn't mean that you can't form the collection of all sets that don't contain themselves and think about it, just like you can form the Cartesian product as the collection of all pairs and think about it and come to the obvious conclusion that it contains a choice function; the paradox arises only if you then insist on calling that collection a "set". So axiomatic set theory, in a sense, allows us to define formally and without many of the pitfalls of "naive set theory" what sort of things we want to call a "set" in a given context -- that's no reason to doubt the existence of other collections, or not to call these collections "sets" in a different context (if that happens to be consistent).

joriki
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This is only half an answer, since I don't know much about the axiom of determinacy and so cannot explain how it conflicts with the axiom of choice. However, I hope that the exposition below is useful in answering your first question.


Axiom of choice. Let $\{ X_i : i \in I \}$ be a set, with each $X_i$ a non-empty set. Let $X = \bigcup_{i \in I} X_i$, the union of all of them. Then, there is a function $f : I \to X$ such that $f(i) \in X_i$ for each $i$ in $I$.

Intuitively, this is saying that if you give me any collection of non-empty sets, I can ‘choose’ an element from each one. This is obviously true in the case where you only give me finitely many sets. But what if you give me infinitely many? Obviously I can choose an element for an arbitrarily large number of them, given enough time, but it is not at all obvious that I can, in one fell swoop, choose an element from all of them at once. If I can, that must be because I can describe, in a finitistic way, a choice procedure which does so: this is exactly what $f : I \to X$ is in the above axiom, and for this reason it is called a choice function.

The way I have described the problem perhaps predisposes you to believe that the axiom of choice is not true. How could it be that there is a choice procedure for any arbitrary problem? The axiom of choice is non-constructive and merely asserts the existence of choice functions. But most mathematicians believe that the axiom of choice is true, because it is convenient. Indeed, the axiom of choice is equivalent to many other statements, some more obviously ‘correct’ or ‘wrong’ than others. A famous joke, apparently due to Jerry Bona, references this:

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

I'll start with some of the less set-theoretic ones, to show how it connects with ‘real’ mathematics.

Equivalent 1. Every non-trivial vector space has at least one basis.

Like the axiom of choice, this seems entirely reasonable in finite settings: for example, a finite-dimensional vector space obviously has a basis (but how would we know it's finite dimensional otherwise?), and similarly, given a finite collection of non-empty sets, it is obviously possible to pick one element from each set.

Equivalent 2. (Krull's theorem) Every non-trivial ring (with identity) has a maximal ideal. This is proven from the axiom of choice using Zorn's lemma.

Equivalent 3. Every surjective function has a right inverse, that is, if $p : X \to Y$ is a surjective function, then there is a function $f : Y \to X$ such that $f(p(y)) = y$ for all $y$ in $Y$. [The function $p$ is surjective just if for every $y$ in $Y$, there is an $x$ in $X$ such that $p(x) = y$.]

It's not hard to see intuitively that this is equivalent to the axiom of choice: after all, what's happening here is that $p$ is partitioning $Y$ into subsets indexed by $X$, and to construct $f$ one needs to pick from each subset one element; conversely, assuming that this is true, given a choice problem, we can reformulate it so as to obtain a solution using this $f$: for convenience, assume that $X_i$ and $X_j$ have no common elements when $i \ne j$, take $X$ as in the axiom, and set $Y = I$, and let $p : X \to I$ be the function which sends $x$ to $i$ if $x \in X_i$. Then any right inverse $f$ is obviously a choice function.

Just to show why this sort of thing might be a little suspect, let's make a tiny tweak to the conditions. Suppose instead that $X$ and $Y$ are topological spaces, and that $p$ is continuous. Then by no means is it guaranteed that there is a continuous right inverse $f$. For example: let $X$ be the real line and $Y$ be the circle. It's obvious that you can wrap $X$ around $Y$ in a continuous way—so there is such a continuous surjective function $p$. But there's no way to take a circle and transform it into a line continuously, so there isn't any continuous right inverse $f$.

Equivalent 4. (Well-ordering principle) Every non-empty set admits a well-ordering. [A well-ordering is an ordering such that every non-empty subset has a unique minimal element.]

Its equivalence with the axiom of choice is also intuitive: if every non-empty set has a well-ordering, then I can just give $I$ a well-ordering, and work through each $X_i$ in order, giving each one a well-ordering and picking a minimal element; conversely, given the axiom of choice, we can just ‘choose’ a well-ordering.

Equivalent 5. Given any two sets $X$ and $Y$, there is either an injection $X \to Y$, or an injection $Y \to X$, or a bijection between the two sets. This basically says that any two cardinal numbers can be compared. [An injection is a function such that different points in the domain get sent to different points in the image, that is, a function $f : X \to Y$ is injective just if $f(x) = f(x')$ implies $x = x'$. A bijection is a function which is both injective and surjective.]

Equivalent 6. Given a family of non-empty sets $\{ X_i : i \in I \}$, their cartesian product $\prod_{i \in I} X_i$ is non-empty. This is really just a reformulation of the axiom of choice, since an element of the cartesian product is precisely a choice function. [The cartesian product of a finite number of sets can be thought of as a set of ordered lists: for example, $X_1 \times X_2$ is the set of pairs $(x_1, x_2)$ where $x_1 \in X_1$ and $x_2 \in X_2$. But such a pair is obviously the same thing as a function from $\{ 1, 2 \}$ to $X_1 \cup X_2$, so that is how we define the cartesian product in the infinite case.]

Equivalent 7. (Tychonoff's theorem) Given a family of (non-empty) compact topological spaces, their cartesian product, equipped with the product topology, is (non-empty and) compact. This is a somewhat advanced bit of mathematics, but I state it here since it is related to the previous example.

Equivalent 8. (Zorn's lemma) Every partially-ordered set with the property that every chain has an upper bound in fact has a maximal element. [A partially-ordered set is a set with a binary relation $\le$ such that $x \le x$ ($\le$ is reflexive); $x \le y$ and $y \le z$ implies $x \le z$ ($\le$ is transitive); and $x \le y$ and $y \le x$ implies $x = y$ ($\le$ is antisymmetric). A chain is a subset which has the additional property that for all $x$ and $y$ in the subset, either $x \le y$ or $y \le x$. A chain $C$ in a partially-ordered set $X$ has an upper bound if there is an element $u$ in $X$ such that $x \le u$ for every $x$ in $C$. A maximal element is an element $m$ such that there do not exist any elements $x$ such that $m \le x$.]


There are also various results which are known to be invalid without the axiom of choice. In other words, these are statements implied by the axiom of choice but which do not themselves imply the axiom of choice, unlike the equivalents listed above. Some of these results are ‘useful’, and others are unpleasant and run counter to intuition.

Consequence 1. Any two bases of a vector space have the same cardinality—in other words, the dimension of a vector space is well-defined and does not depend on the basis. It can be shown (assuming the consistency of certain logical theories) that there is a universe of Zermelo–Fraenkel set theory where the axiom of choice is false (or, for short, ‘a universe without AC’) and in which there is a vector space which have two bases which have different cardinalities.

Consequence 2. Every field has an algebraic closure. [A field $K$ is algebraically closed if every polynomial with coefficients in $K$ also has a root in $K$. The field of complex numbers $\mathbb{C}$ is algebraically closed, for example. An algebraic closure of a field $F$ is a field $K$ which is algebraically closed and contains $F$ as a subfield.] As above, there is a universe without AC which has a field with no algebraic closure.

Consequence 3. Every countable union of countable sets is again countable. [A set $X$ is countable if there is an injection $X \to \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers.] There is a universe without AC in which the real numbers is a countable union of countable sets, but it can be shown that in any universe which obeys ZF, the real numbers must be uncountable.

Consequence 4. A function is sequentially continuous at a point if and only if it is also continuous at that point. [A function $f$ is sequentially continuous just if for every sequence of points $(x_n)$ converging to $x$, $f(x_n)$ converges to $f(x)$. By continuous here we mean the usual $\epsilon$-$\delta$ definition.] But there is a universe without AC in which there is a function of the real numbers which is sequentially continuous at a point but not continuous there.

Consequence 5. (Vitali set) There is a non-Lebesgue-measurable subset of the real line. [Informally, there is a subset of the real line for which does not admit a meaningful definition of ‘total length’ compatible with the intuitive notion that the unit interval $[0, 1]$ has length $1$.] There is a universe without AC in which every subset of Euclidean space is measurable.

Consequence 6. (Banach–Tarski theorem) There is a partition of the ball into finitely many pieces such that by rigid motions, these pieces can be reassembled into two balls of the same volume as the original. These pieces are, of course, not measurable, so do not exist in the universe mentioned above.


But is the axiom of choice true? That's not really a meaningful question. At best we can only ask whether Zermelo–Fraenkel set theory together with the axiom of choice (or ZFC for short) is logically consistent or not. Unfortunately, the answer to this question is as yet unknown, and quite possibly can never be known. (Gödel's incompleteness theorem is a precise form of the latter claim: it shows that the logical consistency of ZFC cannot be proven from the axioms of ZFC alone.) Personally, I advocate for pluralism in mathematics, so I don't see this as much of a problem. If it turns out that Zermelo–Fraenkel set theory with the axiom of choice is consistent and the Zermelo–Fraenkel set theory with some negation of the axiom of choice (e.g. the axiom of determinacy) is consistent, this is all the better, because it adds to the richness of mathematics. It's also a defence against the extremely unlikely event that someone does demonstrate that ZFC is inconsistent, though I suspect most mathematicians would just move on as if nothing ever happened...

Zhen Lin
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  • Are you sure that the axiom of choice follows from the existence of maximal ideals in rings? – Asaf Karagila Jul 20 '11 at 10:59
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    @Asaf: According to Arturo, this was proven by Hodges [1979, MR0533327]. – Zhen Lin Jul 20 '11 at 11:04
  • Interesting. Thanks for the reference! – Asaf Karagila Jul 20 '11 at 11:07
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    Also, if you give important consequences I would add the fact that continuity by $\epsilon-\delta$ is equivalent to continuity by limits. Most people are not aware that this is a consequence of the axiom of choice. – Asaf Karagila Jul 20 '11 at 11:44
  • @Asaf: ... huh, I never thought about that. On that note, do you know of any interesting results whose (in)equivalence with AC is as yet unknown? – Zhen Lin Jul 20 '11 at 12:41
  • There are many many statements which are suspected as equivalent to AC, whose status is unknown. Interesting is a matter of choice, one I think is very interesting is whether or not the statement "Every vector space has a basis" can be restricted to a single field. – Asaf Karagila Jul 20 '11 at 12:55
  • +1 I think your answer is the best so far; however, it leaves me with a question that I originally had: Why, in your example, do you need to chose an element from *multiple* sets? Why not just say, "It's not obvious how you can choose an element from an uncountably infinite set that is not well-ordered."? i.e. Why do you need sets of sets? Isn't also just as impossible to choose a single element from an indistinguishable number of others in the same set? Why do you need other sets? – user541686 Jul 20 '11 at 14:54
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    @Mehrdad: I think that your answer cannot really be answered intuitively, and to have a satisfactory answer you will have to study at least *some* basic axiomatic set theory, logic and how "regular" mathematics can be develop within set theory. Otherwise you will always have more questions to pursue (which is a good thing!) – Asaf Karagila Jul 20 '11 at 15:31
  • *There are also various results which are known to be false without the axiom of choice.* Wouldn't these results then in fact *imply* the AC instead of the other way around? If we have $\neg AC \Rightarrow \neg R$, then $R \Rightarrow AC$. – Paŭlo Ebermann Jul 20 '11 at 15:35
  • @Asaf: As polite as I'd like to be, would you mind stopping intervening into *every* comment I'm placing, and telling me how my question is impossible to answer? If you feel your answer is better, then great! *I got your point already; my question is terrible.* If you'd like to continue the conversation, please try responding to my comment on your own post instead. Thanks. – user541686 Jul 20 '11 at 15:36
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    @Mehrdad: But it is possible to choose an element from a single uncountably infinite set that is not well-ordered, without AC. Indeed, an uncountably infinite set must be non-empty (by definition!), so it has some member, and we are done. The same is true if you give me any *single* non-empty set: once you have told me that the set is non-empty, there's nothing for me to do! And then by induction I can do this when you give me any finite number of non-empty sets. But what if you give me infinitely many non-empty sets? *That* is the puzzle. – Zhen Lin Jul 20 '11 at 15:38
  • @Zhen: Thanks for the explanation! Hm... you mentioned you can prove it with induction? But if you don't know where to start, what would be your base case? I think I'm misunderstanding this, but whether or not a set *has* at least one element and whether or not you can *choose* at least one element are two different things, no? – user541686 Jul 20 '11 at 15:38
  • @Paulo: But how do we know the result is true in the first place? Certainly, if we know a priori that $R$, and if we prove $\lnot \text{AC} \implies \lnot R$, then of course $\text{AC}$ follows. The problem, you see, is that $R$ will have been proven using the axiom of choice, and it is tautological that $\text{AC} \implies \text{AC}$. – Zhen Lin Jul 20 '11 at 15:40
  • @Mehrdad: I admit I was being somewhat informal in my comment. Formally, what we need to show is that a choice function *exists*. But to define this you really have to understand sets of sets, and what it means for a function to exist, and so on. If you're still interested, the induction is on the *number of sets you give me* [not their sizes!], and the base case is the one I described in that comment. – Zhen Lin Jul 20 '11 at 15:43
  • @Zhen: I misunderstood what you were using induction for -- I thought it was for the subsets, not the superset. What I meant to ask was, you say you're "done" when you can prove that there's at least on element in the set, but that doesn't say anything about how to *choose* the element. i.e. How can you choose any element from the set (even if you can prove that there's at least one element in the set) when they have no distinguishing property? (An example might be great.) – user541686 Jul 20 '11 at 15:47
  • @Mehrdad: This is verging on your discussion with Asaf earlier. We use the word ‘choice’ because the concept we are giving a name to is *analogous* to our everyday notion of choice. It is *not* the same thing. To narrow things down a little, the problem is that a choice function can be anything whatsoever. It need not be describable by any finistic means, let alone be computable by an algorithm. At any rate, if you must have an answer: if you give me a set of elements which are all indistinguishable, then it doesn't matter what I choose, because they're all the same! – Zhen Lin Jul 20 '11 at 15:53
  • @Zhen: I see -- I won't discuss the choice further then, but in that case, why can't you say, "If you give me a *lot* of those sets, then it doesn't matter what I choose from each one; they're all the same! So by definition I can choose something from each one."? – user541686 Jul 20 '11 at 15:54
  • @Zhen: My point is that *are known to be false without the axiom of choice* is not the right wording - better would be something like *are known to be false in some models without the axiom of choice*, as there also could be models without the axiom where these results still are true (at least, when they are not equivalent). – Paŭlo Ebermann Jul 20 '11 at 15:57
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    @Mehrdad: Precisely. Another analogy: Imagine your professor gives you a multiple-choice exam with infinitely many questions, each of which has infinitely many options, and tells you that it doesn't matter what answers you submit, just as long as you submit one (and only one!) answer for each question, and you must submit by all your answers by midnight. Obviously, that can't be done... by a human. In this analogy, the axiom of choice is a black box which can meet your professor's demands and instantly produce a completed answer sheet, *by magic*. – Zhen Lin Jul 20 '11 at 16:07
  • @Zhen: Even if he gives only two options for each question, you would still need the axiom of choice... :-) (Also, I like the magic analogy ;-)) – Asaf Karagila Jul 20 '11 at 16:13
  • @Zhen: I think I understand the axiom, but not why it is necessary. i.e. What I'm not understanding is: why *don't* you need the AC for choosing an element from the *subset*, when you *do* need it for repeating that procedure over an infinite number of such subsets? Does *repetition* need an axiom? – user541686 Jul 20 '11 at 16:14
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    @Mehrdad: A useful catchphrase to remember: **Induction does not reach infinity.** Here is an example which should make it obvious. Let $X$ be a set. If $X$ has $0$ members, it is a finite set. Suppose a set with $n$ members is a finite set. Then clearly a set with $n + 1$ members is also a finite set. By induction on the number of members of $X$, every set is finite...! Somehow we've reached the wrong conclusion, and the mistake is in assuming that what is true for arbitrarily large numbers is also true for the infinite. – Zhen Lin Jul 20 '11 at 16:18
  • @Zhen: I *think* I finally understand it now, thanks a lot for explaining that! :) – user541686 Jul 20 '11 at 16:20
  • @Zhen: Unfortunately I'm unaccepting -- I totally forgot you didn't answer the other part of the question! :( Still +1, though! – user541686 Jul 20 '11 at 17:41
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    @Zhen: Re-reading your answer, the last paragraph is somewhat ambiguous. Godel proved that if ZF is consistent then ZFC is consistent and Paul Cohen proved that if ZFC is consistent then ZF+~AC is consistent as well. It is also very easy to have the consistency strength of fragments of AC. Determinacy is a much stronger axiom than a form of negation of AC. It is equiconsistent to the existence of some large cardinals, which in turn implies a very strong form of consistency of ZF. That is Con(ZF+AD)->Con(ZF), while Con(ZF)<->Con(ZFC)<->Con(ZF+~AC). – Asaf Karagila Jul 26 '11 at 10:23
  • Did you mean f(p(x))=x for equivalent 3? – Casebash Nov 25 '11 at 13:55
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I'm late to the party but I just read the following explanation of the Axiom of Choice and thought it was good, so I'm adding it. (Quoted from Just / Weese, p. 118):


(A9)Axiom of Choice: For every family $x$ of non-empty pairwise disjoint sets there exists a set $z$ such that $|z \cap y|=1$ for each $y \in x$.

The Axion of Choice generalises the familiar process of choosing representatives. If $x$ is the family of congressional districts, then the Congress of the United States is an example of a set $z$as in (A9). Note that congressional districts are non-empty and disjoint.

What makes the Axiom of Choice so special that we even mention it specifically in the name of our theory ZFC? Its non-constructive character. The axiom assures us that there will be a Congress after the next Election Day but it does not specify who will be elected. Otherwise, there would be no point in holding elections in the first place. The latter option has always been preferred by dictators who evidently do not have much confidence in the Axiom of Choice and prefer legislatures composed of loyal supporters.

Now suppose for the sake of argument that you seized dictatorial power of the United Sets of the Universe. You need to stay on good terms with your archrival, the Former Solid Intersection who keeps meddling in your internal affairs with evil propaganda about human rights, democracy and similar nuisances. In order to keep up appearances you need some sort of congress. So you start looking at the list of congressional districts and make your choices. But there is a problem: The United Sets of the Universe have infinitely many congressional districts. While you are busy picking your best congress, your enemies may be plotting your overturn. You better get quickly done with this congress-choosing business. In other words, you must make infinitely many choices in a finite amount of time without having the option of describing all these choices by a single formula. The Axiom of Choice says that you can somehow do this, but it does not give you much control over the outcome of the selection process.

Rudy the Reindeer
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