Introduction to Mathematical Philosophy

number of collections of sub-classes of is a reflexive number, i.e. that, if is the number of members of , so that is the number of sub-classes of and is the number of collections of sub-classes, then, provided is not inductive, must be reflexive. But this is a long way from what we set out to prove. In order to advance beyond this point, we must employ the multiplicative axiom. From each set of sub-classes let us choose out one, omitting the sub-class consisting of the null-class alone. That is to say, we select one sub-class containing one term, , say; one containing two terms, , say; one containing three, , say; and so on. (We can do this if the multiplicative axiom is assumed; otherwise, we do not know whether we can always do it or not.) We have now a progression , , , ... sub-classes of , instead of a progression of collections of sub-classes; thus we are one step nearer to our goal. We now know that, assuming the multiplicative axiom, if is a non-inductive number, must be a reflexive number. The next step is to notice that, although we cannot be sure that new members of come in at any one specified stage in the progression , , , ... we can be sure that new members keep on coming in from time to time. Let us illustrate. The class , which consists of one term, is a new beginning; let the one term be . The class , consisting of two terms, may or may not contain ; if it does, it introduces one new term; and if it does not, it must introduce two new terms, say , . In this case it is possible that consists of , , , and so introduces no new terms, but in that case must introduce a new term. The first classes , , , ... contain, at the very most, terms, i.e. terms; thus it would be possible, if there were no repetitions in the first classes, to go on with only repetitions from the [Pg 128] class to the class. But by that time the old terms would no longer be sufficiently numerous to form a next class with the right number of members, i.e. , therefore new terms must come in at this point if not sooner. It follows that, if we omit from our progression , , ,... all those classes that are composed entirely of members that have occurred in previous classes, we shall still have a progression. Let our new progression be called , , .... (We shall have and , because and must introduce new terms. We may or may not have , but, speaking generally, will be , where is some number greater than ; i.e. the 's are some of the 's.) Now these 's are such that any one of them, say , contains members which have not occurred in any of the previous 's. Let be the part of which consists of new members. Thus we get a new progression , , ,... (Again will be identical with and with ; if does not contain the one member of , we shall have , but if does contain this one member, will consist of the other member of ). This new progression of 's consists of mutually exclusive classes. Hence a selection from them will be a progression; i.e. if is the member of , is a member of , is a member of , and so on; then , , , ... is a progression, and is a sub-class of . Assuming the multiplicative axiom, such a selection can be made. Thus by twice using this axiom we can prove that, if the axiom is true, every non-inductive cardinal must be reflexive. This could also be deduced from Zermelo's theorem, that, if the axiom is true, every class can be well ordered; for a well-ordered series must have either a finite or a reflexive number of terms in its field. There is one advantage in the above direct argument, as against deduction from Zermelo's theorem, that the above argument does not demand the universal truth of the multiplicative axiom, but only its truth as applied to a set of classes. It may happen that the axiom holds for classes, though not for larger numbers of classes. For this reason it is better, when [Pg 129] it is possible, to content ourselves with the more restricted assumption. The assumption made in the above direct argument is that a product of factors is never zero unless one of the factors is zero. We may state this assumption in the form: " is a multipliable number," where a number is defined as "multipliable" when a product of factors is never zero unless one of the factors is zero. We can prove that a finite number is always multipliable, but we cannot prove that any infinite number is so. The multiplicative axiom is equivalent to the assumption that all cardinal numbers are multipliable. But in order to identify the reflexive with the non-inductive, or to deal with the problem of the boots and socks, or to show that any progression of numbers of the second class is of the second class, we only need the very much smaller assumption that is multipliable. It is not improbable that there is much to be discovered in regard to the topics discussed in the present chapter. Cases may be found where propositions which seem to involve the multiplicative axiom can be proved without it. It is conceivable that the multiplicative axiom in its general form may be shown to be false. From this point of view, Zermelo's theorem offers the best hope: the continuum or some still more dense series might be proved to be incapable of having its terms well ordered, which would prove the multiplicative axiom false, in virtue of Zermelo's theorem. But so far, no method of obtaining such results has been discovered, and the subject remains wrapped in obscurity. [Pg 130]