element of f(n) for each n. Define S = {f(n): n a natural number}, the range of f, which can be seen to be a set from the axiom schema of replacement. Applying the axiom of regularity to S, let B be an element of S which is disjoint from S. By the definition of S, B must be f(k) for some natural number k. However, we are given that f(k) contains f(k+1) which is also an element of S. So f(k+1) is in the intersection of f(k) and S. This contradicts the fact that they are disjoint sets. Since our supposition led to a contradiction, there must not be any such function, f. The nonexistence of a set containing itself can be seen as a special case where the sequence is infinite and constant. Notice that this argument only applies to functions f that can be represented as sets as opposed to undefinable classes. The hereditarily finite sets, Vω, satisfy the axiom of regularity (and all other axioms of ZFC except the axiom of infinity). So if one forms a non-trivial ultrapower of Vω, then it will also satisfy the axiom of regularity. The resulting model will contain elements, called non-standard natural numbers, that satisfy the definition of natural numbers in that model but are not really natural numbers. They are "fake" natural numbers which are "larger" than any actual natural number. This model will contain infinite descending sequences of elements. For example, suppose n is a non-standard natural number, then and , and so on. For any actual natural number k, . This is an unending descending sequence of elements. But this sequence is not definable in the model and thus not a set. So no contradiction to regularity can be proved. Simpler set-theoretic definition of the ordered pair The axiom of regularity enables defining the ordered pair (a,b) as {a,{a,b}}; see ordered pair for specifics. This definition eliminates one pair of braces from the canonical Kuratowski definition (a,b) = {{a},{a,b}}. Every set has an ordinal rank This was actually the original form of the axiom in von Neumann's axiomatization. Suppose x is any set. Let t be the transitive closure of {x}. Let u be the subset of t consisting of unranked sets. If u is empty, then x is ranked and we are done. Otherwise, apply the axiom of regularity to u to get an element