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Transcript
The Axiom of Choice If you read about set theory, or take more advanced math classes, one thing you will almost certainly hear about is the axiom of choice. I thought you might like to hear a little bit about the axiom of choice, and related subjects. As you probably know, it is possible to make discussions involving sets precise via “axiomatic set theory.” In this class, we have chosen a more informal approach to sets, simply thinking of them as bins that we throw objects (numbers, other sets, or whatever) into. In fact, this is how the vast majority of mathematicians think about sets. It is rare for mathematicians to make any direct mention of set-theoretic axioms, with one exception: the axiom of choice. Proofs which make use of the axiom of choice have a very different flavor from those that do not, and there is a feeling that using the axiom of choice is legitimate (if it is needed), but not very satisfying. Therefore, authors using axiom of choice often say directly that they are using it. So what is the axiom of choice? Basically, it is the following very believable idea: if I have a set X whose elements are nonempty sets, then then I can choose one element from each of the sets in X. You may be surprised that we need an axiom for this! It seems downright obvious. But it’s a little bit like how we needed an axiom to justify proof by induction. Recall that in that case, if we had any finite number of cases to check, and we can prove that the first is true and each one implies the next, then we will have that all of them are true, without any “axiom of induction.” The trouble comes when we try to do this for an infinite number of cases: a “proof” (by definition) is only allowed to have finitely many steps. We have that case 1 is true, and since 1 =⇒ 2 case 2 is true, and similarly case 3 is true, and case 4 is true; but to conclude that all cases n ≥ 1 are true requires the induction axiom. Similarly, if X were any finite set consisting of nonempty sets, then there would be no problem in choosing one element from each of the members of X. The trouble is that when X has infinitely many members, how do we know that we can make all of the choices in a “finite amount of time”? So we need an axiom for this. Interestingly, there are several statements that turn out to be logically equivalent to the axiom of choice, although they do not appear to be the same at all. I think this is partly why the axiom of choice shows up so often. Let me mention a couple of these which have to do with order relations. One statement that’s equivalent to the axiom of choice is the well-ordering theorem. It says that for any set A, there exists a total order ≤ on A such that for any nonempty subset K ⊆ A, K has a minimum element. Such a total order is called a well-order. For example, the standard order on the natural numbers N is a well-order. The standard 1 order on the integers Z is not a well-ordering: choose any subset which is not bounded below, e.g. {0, −1, −2, . . .} (or for that matter all of Z), and such a subset will not have a least element. However, you can come up with a different total ordering of the integers which is a well-ordering; for example, define ≤ by 0 ≤ 1 ≤ −1 ≤ 2 ≤ −2 ≤ . . .. That was easy, but note that it is not at all clear we can define a well-order on the real numbers. To do that, it turns out that we will need something like the axiom of choice. At this point, it may be possible to start to give a sense of why things that use the axiom of choice feel different from things that do not. We did not need to use any mumbo jumbo to define a well-ordering on the integers; we simply wrote one down explicitly. For the real numbers, we were not able to do this. We can show, using the axiom of choice, that there exists a well-ordering on the real numbers, but we will never be able to give any explicit description of what it is (this can be proven). Statements that “there exists ...”, without saying (or being able to say) what it is, abound in modern mathematics, and often involve use of the axiom of choice. This bothers “constructivists,” mathematicians who believe that to prove the existence of something one must explicitly write down what it is. Most mathematicians today are willing to put up with non-constructive existence theorems, but prefer constructive ones when possible. Another statement that is equivalent to the axiom of choice is called Zorn’s lemma. This is the version of the axiom of choice I have seen by far the most often in my studies. In order to understand it, let’s define the notion of a “chain.” Let (A, ≤) be a partially ordered set. We define a chain in A to be a totally-ordered subset of A. That is, we call K ⊆ A a chain if for every x, y ∈ K, we have either x ≤ y or y ≤ x. Then Zorn’s lemma says this: if (A, ≤) is a partially ordered set such that every chain has an upper bound, then A must contain a maximal element (possibly more than one maximal element). Confused? You’re not alone. There’s a joke, popular among mathematicians, that goes like this: “The axiom of choice is obviously true, the well-ordering theorem is obviously false, and who can tell about Zorn’s lemma?” The joke is that many mathematicians find the axiom of choice to be intuitive, the well-ordering theorem to be counterintuitive, and Zorn’s lemma to be too complex for any intuition. (See http://en.wikipedia.org/wiki/Axiom_of_choice#Quotes.) Here’s how Zorn’s lemma often comes up: let B be some set. Suppose there is some property P which is satisfied by some, but not all, of the subsets of B. We sometimes want to show that there is a maximal subset satisfying property P : a subset which satisfies property P , and which is not contained in any larger subset that also satisfies property P . Let’s see how one might use Zorn’s lemma to show that there is such a maximal subset. 2 Let A be all the subsets of B satisfying property P (so A is a subset of the power set of B). We’ll define a partial order on A like this: for X, Y ∈ A (i.e. X and Y are subsets of B satisfying property P ), X ≤ Y if and only if X ⊆ Y . This is the standard partial order on the power set of a set (or on a subset of the power set); your book mentions it. Suppose we can show that the union of subsets satisfying P is a subset which also satisfies P . Then we’re in business: suppose {Xi }i∈I (I some index set) is a chain of subsets of B satisfying P (remember, chain means totally-ordered [ subset, so here that means that for any i, j ∈ I, either Xi ⊆ Xj or Xj ⊆ Xi ). Then Xi is a subset of i∈I B satisfying P , and containing all the Xi ’s, i.e. it is an upper bound of the elements Xi ∈ A. Therefore, we can apply Zorn’s lemma – since an arbitrary chain in A had an upper bound, A must have a maximal element! That is to say, a subset of B satisfying P such that there is no subset containing it that satisfies P , which is exactly what we were seeking. Does it seem somehow unsatisfying that we magically showed that there has to be a maximal subset satisfying P , without giving any indication of what it might be? Again, this is a typical example of a proof using the axiom of choice (in Zorn’s lemma form). Zorn’s lemma can also be used to prove statements like this: Every vector space has a basis (specifically, a “Hamel” basis). See http://planetmath.org/encyclopedia/ EveryVectorSpaceHasABasis.html. You may have seen that this is an easy fact for finite-dimensional vector spaces, but it’s not even true for infinite-dimensional vector spaces, unless we assume the axiom of choice (or equivalently, Zorn’s lemma). Somehow, we would like the statement that every vector space has a basis to be true. We know it’s true for finite-dimensional vector spaces, and the whole thing feels neater if it’s true in general. So by assuming the axiom of choice, we can prove that some things behave as they “should” in general. But, the axiom of choice can also be used to prove some very bizarre things. For example, one can use it to prove that a solid ball in 3-dimensional space can be split into a finite number of non-overlapping pieces, which can then be put back together in a different way to yield two identical copies of the original ball. The reassembly process involves only moving the pieces around and rotating them, without changing their shape. See http://en.wikipedia.org/wiki/Banach-Tarski. So, is the axiom of choice true, or untrue? Initially, it met with a lot of resistance among mathematicians. Gradually, it gained acceptance, so that today its use is completely standard, and only a tiny minority object. But this doesn’t really answer the question: is it true or not? My personal opinion is that it’s really meaningless to say whether the axiom of choice is true or not. It is merely an axiom, an assumption. One can look at the mathematics that one gets by assuming it, and the math one gets if 3 one doesn’t assume it, and compare the two, and the differences will be interesting. Neither is “right” or “wrong” in the absolute; they are simply implied by the initial assumptions (axioms). I think an analogy can be made with playing games. Different games have different rules, just as one can explore the consequences of different sets of mathematical axioms. A collection of rules for playing a game can’t be “wrong” (at least, unless the rules contradict each other), but different rules will certainly yield different games. Still, some rules are more basic than others. For instance, in most games it is an unwritten rule that one may not kill opposing players. Similarly, most of math is based in some way on sets, and all of math is based on the basic axioms of logic. Maybe even these axioms are not “true” in any absolute sense. But altering these most basic axioms might result in something too foreign, too distant from intuition for mathematicians to find it interesting. Then again, if the axioms yielded a rich, surprising, beautiful structure, who knows? 4