No set is an element of itself Let A be a set, and apply the axiom of regularity to \{A\}, which is a set by the
axiom of pairing. By this axiom, there must be an element of \{A\} which is disjoint from \{A\}. Since the only element of \{A\} is A, it must be that A is disjoint from \{A\}. So, since A \cap \{A\} = \varnothing, we cannot have A an element of A (by the definition of
disjointness).
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. 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
structure will contain elements, called non-standard natural numbers, that satisfy the definition of natural numbers in that structure but are not really natural numbers. They are "fake" natural numbers that are "larger" than any actual natural number. This structure will contain infinite descending (with respect to the interpretation of \in in the structure) sequences of elements. For example, suppose
n is a non-standard natural number, then (n-1) \in n and (n-2) \in (n-1), and so on. For any actual natural number
k, (n-k-1) \in (n-k). This is an unending descending sequence of elements. But this sequence is not definable in the structure 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. To prove it from the axiom of regularity, 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
w of
u that is disjoint from
u. Since
w is in
u,
w is unranked.
w is a subset of
t by the definition of transitive closure. Since
w is disjoint from
u, every element of
w is ranked. Applying the axioms of replacement and
union to combine the ranks of the elements of
w, we get an ordinal rank for
w, to wit \textstyle \operatorname{rank} (w) = \cup \{ \operatorname{rank} (z) + 1 \mid z \in w \}. This contradicts the conclusion that
w is unranked. So the assumption that
u was non-empty must be false and
x must have rank.
For every two sets, only one can be an element of the other Let
X and
Y be sets. Then apply the axiom of regularity to the set {
X,
Y} (which exists by the axiom of pairing). We see there must be an element of {
X,
Y} that is also disjoint from it. It must be either
X or
Y. By the definition of disjoint then, we must have either
Y is not an element of
X or vice versa. ==The axiom of dependent choice and no infinite descending sequence of sets implies regularity==