In the first place, it is clear that the number of sub-classes of a given class (say ) is at least as great as the number of members, since each member constitutes a sub-class, and we thus have a correlation of all the members with some of the sub-classes. Hence it follows that, if the number of sub-classes is not equal to the number of members, it must be greater. Now it is easy to prove that the number is not equal, by showing that, given any one-one relation whose domain is the members and whose converse domain is contained among the set of sub-classes, there must be at least one sub-class not belonging to the converse domain. The proof is as follows:[21] When a one-one correlation is established between all the members of and some of the sub-classes, it may happen that a given member is correlated with a sub-class of which it is a member; or, again, it may happen that is correlated with a sub-class of which it is not a member. Let us form the whole class, say, of those members which are correlated with sub-classes of which they are not members. This is a sub-class of , and it is not correlated with any member of . For, taking first the members of , each of them is (by the definition of ) correlated with some sub-class of which it is not a member, and is therefore not correlated with . Taking next the terms which are not members of , each of them (by the definition of ) is correlated with some sub-class of which it is a member, and therefore again is not correlated with . Thus no member of is correlated with . Since was any one-one correlation of all members [Pg 85] with some sub-classes, it follows that there is no correlation of all members with all sub-classes. It does not matter to the proof if has no members: all that happens in that case is that the sub-class which is shown to be omitted is the null-class. Hence in any case the number of sub-classes is not equal to the number of members, and therefore, by what was said earlier, it is greater. Combining this with the proposition that, if is the number of members, is the number of sub-classes, we have the theorem that is always greater than , even when is infinite. [21]This proof is taken from Cantor, with some simplifications: see Jahresbericht der deutschen Mathematiker-Vereinigung, I. (1892), p. 77. It follows from this proposition that there is no maximum to the infinite cardinal numbers. However great an infinite number may be, will be still greater. The arithmetic of infinite numbers is somewhat surprising until one becomes accustomed to it. We have, for example, (This follows from the case of the ratios, for, since a ratio is determined by a pair of inductive numbers, it is easy to see that the number of ratios is the square of the number of inductive numbers, i.e. it is ; but we saw that it is also .)