Definitions

# Axiom of regularity

In mathematics, the axiom of regularity (also known as the axiom of foundation) is one of the axioms of Zermelo-Fraenkel set theory and was introduced by . In first-order logic the axiom reads:

$forall A \left(exists B \left(B in A\right) rightarrow exists B \left(B in A land lnot exist C \left(C in A land C in B\right)\right)\right).$

Or in prose:

Every non-empty set A contains an element B which is disjoint from A.

Two results which follow from the axiom are that "no set is an element of itself", and that "there is no infinite sequence (an) such that ai+1 is an element of ai for all i".

With the axiom of choice, this result can be reversed: if there are no such infinite sequences, then the axiom of regularity is true. Hence, the axiom of regularity is equivalent, given the axiom of choice, to the alternative axiom that there are no downward infinite membership chains.

The axiom of regularity is arguably the least useful ingredient of Zermelo-Fraenkel set theory, since virtually all results in the branches of mathematics based on set theory hold even in the absence of regularity (see Chapter III of [Kunen 1980]). However, it is used extensively in establishing results about well-ordering and the ordinals in general. In addition to omitting the axiom of regularity, non-standard set theories have indeed postulated the existence of sets that are elements of themselves.

Given the other ZFC axioms, the axiom of regularity is equivalent to the axiom of induction.

## Elementary implications of Regularity

### No set is an element of itself

Let A be a set such that A is an element of itself and define B = {A}, which is a set by the axiom of pairing. Applying the axiom of regularity to B, we see that the only element of B, namely, A, must be disjoint from B. But A is both an element of itself and an element of B. Thus B does not satisfy the axiom of regularity and we have a contradiction, proving that A cannot exist.

### No infinite descending sequence of sets exists

Suppose, to the contrary, that there is a function, f, on the natural numbers with f(n+1) an 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.

Notice that this argument only applies to functions f which 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, which 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 $\left(n-1\right) in n$ and $\left(n-2\right) in \left(n-1\right)$, and so on. For any actual natural number k, $\left(n-k-1\right) in \left(n-k\right)$. 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.

### The axiom of choice and no infinite descending sequence of sets implies Regularity

Let the non-empty set S be a counter-example to the axiom of regularity; that is, every element of S has a non-empty intersection with S. Let g be a choice function for S, that is, a map such that g(s) is an element of s for each non-empty subset s of S. Now define the function f on the non-negative integers recursively as follows:

$f\left(0\right) = g\left(S\right),!$
$f\left(n+1\right) = g\left(f\left(n\right) cap S\right).,!$

Then for each n, f(n) is an element of S and so its intersection with S is non-empty, so f(n+1) is well-defined and is an element of f(n). So f is an infinite descending chain. This is a contradiction, hence no such S exists.

### 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}}.

## Regularity does not resolve Russell's paradox

In naive set theory, Russell's paradox is the fact "the set of all sets that do not contain themselves as members" leads to a contradiction. The paradox shows that that set cannot be constructed using any consistent set of axioms for set theory. Even though the axiom of regularity implies that no set contains itself as a member, that axiom does not banish Russell's paradox from Zermelo-Fraenkel set theory (ZF). In fact, if the ZF axioms without Regularity were already inconsistent, then adding Regularity would not make them consistent. Russell's paradox is impossible in ZF because the sets that ZF's axiom of separation allows must be subsets of some existing set. Hence the collection of all sets which do not contain themselves is not a set but a proper class.

## Regularity and cumulative hierarchy

In ZF it can be proven that the class $bigcup_\left\{alpha\right\} V_alpha !$ (see cumulative hierarchy) is equal to the class of all sets. This statement is even equivalent to the axiom of regularity (if we work in ZF with this axiom omitted). From any model which does not satisfy axiom of regularity, a model which satisfies it can be constructed by taking only sets in $bigcup_\left\{alpha\right\} V_alpha !$.