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.