Introduction to Mathematical Philosophy

is at least one one-many relation implying and having for its converse domain. The multiplicative axiom is equivalent to the assumption that if be any class, and all the sub-classes of with the exception [Pg 122] of the null-class, then there is at least one selector from . This is the form in which the axiom was first brought to the notice of the learned world by Zermelo, in his "Beweis, dass jede Menge wohlgeordnet werden kann."[25] Zermelo regards the axiom as an unquestionable truth. It must be confessed that, until he made it explicit, mathematicians had used it without a qualm; but it would seem that they had done so unconsciously. And the credit due to Zermelo for having made it explicit is entirely independent of the question whether it is true or false. [25]Mathematische Annalen, vol. LIX. pp. 514-6. In this form we shall speak of it as Zermelo's axiom. The multiplicative axiom has been shown by Zermelo, in the above-mentioned proof, to be equivalent to the proposition that every class can be well-ordered, i.e. can be arranged in a series in which every sub-class has a first term (except, of course, the null-class). The full proof of this proposition is difficult, but it is not difficult to see the general principle upon which it proceeds. It uses the form which we call "Zermelo's axiom," i.e. it assumes that, given any class , there is at least one one-many relation whose converse domain consists of all existent sub-classes of and which is such that, if has the relation to , then is a member of . Such a relation picks out a "representative" from each sub-class; of course, it will often happen that two sub-classes have the same representative. What Zermelo does, in effect, is to count off the members of , one by one, by means of and transfinite induction. We put first the representative of ; call it . Then take the representative of the class consisting of all of except ; call it . It must be different from , because every representative is a member of its class, and is shut out from this class. Proceed similarly to take away , and let be the representative of what is left. In this way we first obtain a progression , , ... , ..., assuming that is not finite. We then take away the whole progression; let be the representative of what is left of . In this way we can go on until nothing is left. The successive representatives will form a [Pg 123] well-ordered series containing all the members of . (The above is, of course, only a hint of the general lines of the proof.) This proposition is called "Zermelo's theorem." The multiplicative axiom is also equivalent to the assumption that of any two cardinals which are not equal, one must be the greater. If the axiom is false, there will be cardinals and such that is neither less than, equal to, nor greater than . We have seen that and possibly form an instance of such a pair. Many other forms of the axiom might be given, but the above are the most important of the forms known at present. As to the truth or falsehood of the axiom in any of its forms, nothing is known at present. The propositions that depend upon the axiom, without being known to be equivalent to it, are numerous and important. Take first the connection of addition and multiplication. We naturally think that the sum of mutually exclusive classes, each having terms, must have terms. When is finite, this can be proved. But when is infinite, it cannot be proved without the multiplicative axiom, except where, owing to some special circumstance, the existence of certain selectors can be proved. The way the multiplicative axiom enters in is as follows: Suppose we have two sets of mutually exclusive classes, each having terms, and we wish to prove that the sum of one set has as many terms as the sum of the other. In order to prove this, we must establish a one-one relation. Now, since there are in each case classes, there is some one-one relation between the two sets of classes; but what we want is a one-one relation between their terms. Let us consider some one-one relation between the classes. Then if and are the two sets of classes, and is some member of , there will be a member of which will be the correlate of with respect to . Now and each have terms, and are therefore similar. There are, accordingly, one-one correlations of and . The trouble is that there are so many. In order to obtain a one-one correlation of the sum of with the sum of , we have to pick out one selection from a set of classes [Pg 124] of correlators, one class of the set being all the one-one correlators of with . If and are infinite, we cannot in general know that such a selection exists, unless we can know that the multiplicative axiom is true. Hence we cannot establish the usual kind of connection between addition and multiplication. This fact has various curious consequences. To begin with, we know that . It is commonly inferred from this that the sum of classes each having members must itself have members, but this inference is fallacious, since we do not know that the number of terms in such a sum is , nor consequently that it is . This has a bearing upon the theory of transfinite ordinals. It is easy to prove that an ordinal which has predecessors must be one of what Cantor calls the "second class," i.e. such that a series having this ordinal number will have terms in its field. It is also easy to see that, if we take any progression of ordinals of the second class, the predecessors of their limit form at most the sum of classes each having terms. It is inferred thence—fallaciously, unless the multiplicative axiom is true—that the predecessors of the limit are in number, and therefore that the limit is a number of the "second class." That is to say, it is supposed to be proved that any progression of ordinals of the second class has a limit which is again an ordinal of the second class. This proposition, with the corollary that (the smallest ordinal of the third class) is not the limit of any progression, is involved in most of the recognised theory of ordinals of the second class. In view of the way in which the multiplicative axiom is involved, the proposition and its corollary cannot be regarded as proved. They may be true, or they may not. All that can be said at present is that we do not know. Thus the greater part of the theory of ordinals of the second class must be regarded as unproved. Another illustration may help to make the point clearer. We know that . Hence we might suppose that the sum of pairs must have terms. But this, though we can prove that it is sometimes the case, cannot be proved to happen always [Pg 125] unless we assume the multiplicative axiom. This is illustrated by the millionaire who bought a pair of socks whenever he bought a pair of boots, and never at any other time, and who had such a passion for buying both that at last he had pairs of boots and pairs of socks. The problem is: How many boots had he, and how many socks? One would naturally suppose that he had twice as many boots and twice as many socks as he had pairs of each, and that therefore he had of each, since that number is not increased by doubling. But this is an instance of the difficulty, already noted, of connecting the sum of classes each having terms with . Sometimes this can be done, sometimes it cannot. In our case it can be done with the boots, but not with the socks, except by some very artificial device. The reason for the difference is this: Among boots we can distinguish right and left, and therefore we can make a selection of one out of each pair, namely, we can choose all the right boots or all the left boots; but with socks no such principle of selection suggests itself, and we cannot be sure, unless we assume the multiplicative axiom, that there is any class consisting of one sock out of each pair. Hence the problem. We may put the matter in another way. To prove that a class has terms, it is necessary and sufficient to find some way of arranging its terms in a progression. There is no difficulty in doing this with the boots. The pairs are given as forming an , and therefore as the field of a progression. Within each pair, take the left boot first and the right second, keeping the order of the pairs unchanged; in this way we obtain a progression of all the boots. But with the socks we shall have to choose arbitrarily, with each pair, which to put first; and an infinite number of arbitrary choices is an impossibility. Unless we can find a rule for selecting, i.e. a relation which is a selector, we do not know that a selection is even theoretically possible. Of course, in the case of objects in space, like socks, we always can find some principle of selection. For example, take the centres of mass of the socks: there will be points in space such that, with any [Pg 126] pair, the centres of mass of the two socks are not both at exactly the same distance from ; thus we can choose, from each pair, that sock which has its centre of mass nearer to . But there is no theoretical reason why a method of selection such as this should always be possible, and the case of the socks, with a little goodwill on the part of the reader, may serve to show how a selection might be impossible. It is to be observed that, if it were impossible to select one out of each pair of socks, it would follow that the socks could not be arranged in a progression, and therefore that there were not of them. This case illustrates that, if is an infinite number, one set of pairs may not contain the same number of terms as another set of pairs; for, given pairs of boots, there are certainly boots, but we cannot be sure of this in the case of the socks unless we assume the multiplicative axiom or fall back upon some fortuitous geometrical method of selection such as the above. Another important problem involving the multiplicative axiom is the relation of reflexiveness to non-inductiveness. It will be remembered that in Chapter VIII. we pointed out that a reflexive number must be non-inductive, but that the converse (so far as is known at present) can only be proved if we assume the multiplicative axiom. The way in which this comes about is as follows:— It is easy to prove that a reflexive class is one which contains sub-classes having terms. (The class may, of course, itself have terms.) Thus we have to prove, if we can, that, given any non-inductive class, it is possible to choose a progression out of its terms. Now there is no difficulty in showing that a non-inductive class must contain more terms than any inductive class, or, what comes to the same thing, that if is a non-inductive class and is any inductive number, there are sub-classes of that have terms. Thus we can form sets of finite sub-classes of : First one class having no terms, then classes having 1 term (as many as there are members of ), then classes having [Pg 127] 2 terms, and so on. We thus get a progression of sets of sub-classes, each set consisting of all those that have a certain given finite number of terms. So far we have not used the multiplicative axiom, but we have only proved that the