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.