has greater cardinality than A. This is known as Cantor's theorem. The argument below is a modern version of Cantor's argument that uses power sets (for his original argument, see Cantor's diagonal argument). By presenting a modern argument, it is possible to see which assumptions of axiomatic set theory are used. The first part of the argument proves that N and P(N) have different cardinalities: There exists at least one infinite set. This assumption (not formally specified by Cantor) is captured in formal set theory by the axiom of infinity. This axiom implies that N, the set of all natural numbers, exists. P(N), the set of all subsets of N, exists. In formal set theory, this is implied by the power set axiom, which says that for every set there is a set of all of its subsets. The concept of "having the same number" or "having the same cardinality" can be captured by the idea of one-to-one correspondence. This (purely definitional) assumption is sometimes known as Hume's principle. As Frege said, "If a waiter wishes to be certain of laying exactly as many knives on a table as plates, he has no need to count either of them; all he has to do is to lay immediately to the right of every plate a knife, taking care that every knife on the table lies immediately to the right of a plate. Plates and knives are thus correlated one to one." Sets in such a correlation are called equinumerous, and the correlation is called a one-to-one correspondence. A set cannot be put into one-to-one correspondence with its power set. This implies that N and P(N) have different cardinalities. It depends on very few assumptions of set theory, and, as John P. Mayberry puts it, is a "simple and beautiful argument" that is "pregnant with consequences". Here is the argument: Let be a set and be its power set. The following theorem will be proved: If is a function from to then it is not onto. This theorem implies that there is no one-to-one correspondence between and since such a correspondence must be onto. Proof of theorem: Define the diagonal subset Since proving that for all will imply that is not onto. Let Then which implies So if then and if then Since one of these sets contains and the other does not, Therefore, is not in the image of ,