Editing Axiom of regularity (section)

==Elementary implications of regularity==

===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]]. We see that 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 <math display="inline">A \cap \{A\} = \varnothing</math>, we cannot have ''A''  an element of ''A'' (by the definition of [[Disjoint sets|disjoint]]).

===No infinite descending sequence of sets exists===
Suppose, to the contrary, that there is a [[function (mathematics)|function]], ''f'', on the [[natural number]]s 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 (set theory)|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 set]]s, ''V''<sub>ω</sub>, satisfy the axiom of regularity (and all other axioms of [[ZFC]] except the [[axiom of infinity]]). So if one forms a non-trivial [[ultraproduct|ultrapower]] of V<sub>ω</sub>, then it will also satisfy the axiom of regularity. The resulting [[model (logic)|model]] <!--WHAT 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.{{dubious|date=February 2023|reason=They satisfy the first-order Peano axioms, so it seems dubious to claim that they are not actually natural numbers. They presumably do not satisfy the second-order Peano axioms with respect to the subset relation of the "ambient" set theory inside of which the model is constructed. But don't they actually satisfy the second-order Peano axioms with respect to the internal subset relation of the model?}} They are "fake" natural numbers which are "larger" than any actual natural number. This model will contain infinite descending sequences of elements.{{clarification needed|date=February 2023|reason=Is the set membership relation in this infinite descending chain the "internal" set membership relation of the model? (I.e. the model's interpretation of the set membership relation?) Or is what follows referring to the set membership relation of the "ambient" set theory in which the model is constructed? Presumably it can't be the latter, because the fact that the latter has a von Neumann cumulative hierarchy, e.g. V_omega, seems to presuppose that it satisfies regularity, and thus otherwise this section would be describing a contradiction. If so, then this section ideally would clarify that what follows refers to the model's interpretation of the set membership relation, and that this is necessarily distinct from (in particular not the restriction of) the ambient set theory's set membership relation.}} For example, suppose ''n'' is a non-standard natural number, then <math display="inline">(n-1) \in n</math> and <math display="inline">(n-2) \in (n-1)</math>, and so on. For any actual natural number ''k'', <math display="inline">(n-k-1) \in (n-k)</math>. 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'') = <nowiki>{{</nowiki>''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 (set)|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'' which 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 <math display="inline">\textstyle \operatorname{rank} (w) = \cup \{ \operatorname{rank} (z) + 1 \mid z \in w \}</math>. 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''} which 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.