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\section*{axiom of foundation}

\hypertarget{the_axiom_of_foundation}{}\section*{{The axiom of foundation}}\label{the_axiom_of_foundation}

\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{alternative_formulations}{Alternative formulations}\dotfill \pageref*{alternative_formulations} \linebreak
\noindent\hyperlink{antifoundation}{Anti-foundation}\dotfill \pageref*{antifoundation} \linebreak
\noindent\hyperlink{structural_meaning}{Structural meaning}\dotfill \pageref*{structural_meaning} \linebreak


In material [[set theory]], the \textbf{axiom of foundation}, also called the \textbf{axiom of regularity}, states that the membership relation $\in$ on the proper class of all [[pure set]]s is [[well-founded relation|well-founded]]. In structural set theory, accordingly, one uses well-founded relations in building structural models of well-founded pure sets.

\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

Given a proper class $A$ of [[pure set]]s, suppose that $A$ has the property that, given any pure set $x$,

\begin{displaymath}
\forall t,\; t \in x \Rightarrow t \in A ,
\end{displaymath}
then $x \in A$. Such an $A$ may be called a \emph{membership-inductive} class. Then the \textbf{axiom of foundation} states that the only membership-inductive class of pure sets is the class of all pure sets. In this form, the axiom of foundation is also called \emph{$\in$-induction}.

Although the statement here refers to proper classes, it can also be formulated as an axiom schema that makes no mention of classes.

\hypertarget{alternative_formulations}{}\subsection*{{Alternative formulations}}\label{alternative_formulations}

While the statement above follows how the axiom of foundation is generally \emph{used} ---to prove properties of pure sets by [[transfinite induction]]---, it is complicated. Two alternative formulations are given by the following lemmas:

\begin{lemma}
\label{InfiniteDescent}\hypertarget{InfiniteDescent}{}
The axiom of foundation holds if and only if there exists no infinite descending [[sequence]] $\cdots \in x_2 \in x_1 \in x_0$.

\end{lemma}
\begin{proof}
First, suppose the axiom of foundation holds, and suppose there were such a sequence. Let $A$ be the class of all sets which are not equal to $x_i$ for any $i$. Then for any $x$, if $x$ were equal to some $x_i$, then some $t\in x$ would also be equal to some $x_j$, namely $x_{i+1}$; hence if all $t\in x$ are in $A$, so is $x$. Thus, by the axiom of foundation, all sets are in $A$, a contradiction.

Second, suppose there are no such sequences, and let $A$ be as in the statement of foundation. Suppose that $A$ does not contain all sets. Then there exists an $x_0\notin A$. By hypothesis on $A$, there must exist an $x_1\in x_0$ such that $x_1 \notin A$. And hence an $x_2\in x_1$ such that $x_2 \notin A$, and so on, producing an infinite descending $\in$-sequence in contradiction to our hypothesis; hence $A$ must contain all sets.

\end{proof}
This is essentially a version of Fermat's method of [[infinite descent]] modified to apply to [[transfinite induction]] instead of only to ordinary [[induction]]. It arguably provides the most direct intuitive picture of what the axiom means. Note that as a special case, it implies that no set can contain itself, since then $\cdots \in x\in x\in x$ would be an infinite chain.

\begin{lemma}
\label{MembershipMinimal}\hypertarget{MembershipMinimal}{}
The axiom of foundation holds if and only if every [[inhabited set|inhabited]] pure set $x$ has a member $y \in x$ such that no $t \in x$ satisfies $t \in y$. (Such a $y$ is called a \emph{membership-minimal} element of $x$.)

\end{lemma}
\begin{proof}
First, suppose the axiom of foundation holds, let $x$ be a pure set without a membership-minimal element, and let $A = \{ y | y\notin x \}$. If $z$ is a set such that all $t\in z$ are in $A$, then we cannot have $z\in x$, since then we would have some $t\in z$ with $t\in x$ and hence $t\notin A$; hence $z\in A$. So $A$ satisfies the assumptions of the foundation axiom, and hence $x$ is empty.

Second, suppose that every inhabited pure set has a membership-minimal element, and let $A$ be as in the statement of foundation. Since every pure set is contained in a [[transitive set]], it suffices to show that $A \cap x = x$ for any transitive $x$. Let $y = \{ z\in x \mid z \notin A \}$. If $y$ is inhabited, then it contains a membership-minimal element $z$, i.e. we have $z\in x$, $z\notin A$, and for every $t\in z$ we have $t\notin z$---but $t\in x$ since $x$ is transitive, hence $t\in A$. Thus this $z$ contradicts the assumption on $A$, so $y$ must be empty, as desired.

\end{proof}
This version is favoured by [[classical mathematics|classical]] set theorists as a statement of the axiom, since it uses neither higher-order reasoning (as our first definition does) or infinity (as infinite descent does).

However, neither of these is acceptable in [[constructive mathematics]], since both lemmas require at least the principle of [[excluded middle]] to prove at least one direction. In particular, the nonexistence of infinite descending sequences is too weak to allow proofs by transfinite induction (except for special forms of $A$), while the requirement that every inhabited pure set have a minimal element is unnecessarily strong and itself implies excluded middle.

More precisely, the necessary set-theoretic axioms for the above proofs are the following. (Is it known, in any case, that proofs using fewer axioms don't exist?)

\begin{itemize}%
\item The ``only if'' direction of Lemma \ref{InfiniteDescent} requires only the axiom of infinity (for ``infinite sequence'' to make sense).


\item The ``if'' direction of Lemma \ref{InfiniteDescent} evidently requires the principle of excluded middle, the axiom of infinity, and the axiom of [[dependent choice]]. It also appears to require the [[axiom of collection]] (since dependent choice, as usually stated, only chooses elements from a sequence of nonempty \emph{sets}, rather than nonempty classes), and a principle of [[induction]] strong enough to recursively construct functions into a [[proper class]] (which is usually proven using the [[axiom of separation]]) in order to put together these nonempty sets into a sequence we can apply dependent choice to.


\item The ``only if'' direction of Lemma \ref{MembershipMinimal} requires only excluded middle.


\item The ``if'' direction of Lemma \ref{MembershipMinimal} requires excluded middle, also that every set is contained in a transitive one (the [[axiom of transitive closure]], which follows from [[axiom of replacement|replacement]]), as well as the [[axiom of separation]] in the form ``the intersection of any class with a set is a set.''



\end{itemize}
Another version of the axiom of foundation which \emph{is} intuitionistically acceptable, but makes no reference to proper classes is:

\begin{lemma}
\label{WFTransitive}\hypertarget{WFTransitive}{}
The axiom of foundation holds if and only if the relation $\in$ on any [[transitive set]] is a [[well-founded relation]].

\end{lemma}
\begin{proof}
Suppose the axiom of foundation holds, let $x$ be a transitive set, and let $S\subseteq x$ be such that for any $y\in x$, if all $t\in y$ are in $S$, then $y\in S$. Let $A$ be the class of all sets $y$ such that if $y\in x$, then $y\in S$; then $A$ satisfies the conditions in the axiom of foundation, so it contains all sets, and hence $S=x$.

Now suppose $\in$ is well-founded on any transitive set and let $A$ satisfy the conditions in foundation. Since every set is contained in a transitive one, it suffices to show that $A\cap x = x$ for any transitive $x$, but this follows directly from the assumption.

\end{proof}
In this case:

\begin{itemize}%
\item The ``only if'' direction requires no notable axioms at all, while


\item The ``if'' direction requires [[transitive closure]], as in Lemma \ref{MembershipMinimal}, and also the axiom of separation.



\end{itemize}
Finally, another commonly cited version of foundation, equivalent to it at least over the other axioms of [[ZF]], is:

\begin{lemma}
\label{CumulativeHierarchy}\hypertarget{CumulativeHierarchy}{}
The axiom of foundation holds if and only if every pure set is an element of $V_\alpha$ for some [[ordinal]] $\alpha$.

\end{lemma}
Here the $V_\alpha$ are the [[cumulative hierarchy]] defined by [[transfinite recursion]] as $V_\alpha = P(\bigcup_{\beta\lt \alpha} V_\beta)$.

\hypertarget{antifoundation}{}\subsection*{{Anti-foundation}}\label{antifoundation}

Most of set theory works without the axiom of foundation, but not the deep study of well-founded pure sets. However, one might want to do material set theory without assuming that all sets are well-founded, then one would not assume this axiom.

Alternatively, one can adopt the \textbf{axiom of anti-foundation}, which says:

\begin{itemize}%
\item Given any [[extensional relation|extensional]] [[binary relation]] $\prec$ on any [[set]] $S$, there exists a unique [[transitive set]] $U$ such that $(U,\in)$ is [[isomorphism|isomorphic]] (necessarily uniquely) to $(S,\prec)$.

\end{itemize}
Since any relation has an [[extensional quotient]], we may also phrase the axiom thus:

\begin{itemize}%
\item Given any binary relation $\prec$ on any [[set]] $S$, there exists a unique [[transitive set]] $U$ and [[surjection]] $f : S \to U$ such that $f(s_1) \in f(s_2)$ if and only if $s_1 \in s_2$, for $s_1, s_2$ in $S$. (That is, $f$ is almost an isomorphism between $(S, \prec)$ and $(U, \in)$, but needn't be [[injection|injective]].)

\end{itemize}
Just as there are several versions of an [[extensional relation]], there are several versions of this axiom. Note that the existence part of the statement is a set-formation axiom, while the uniqueness part is a strong version of the [[axiom of extensionality]] (which is equivalent to the usual one for well-founded sets).

If you include the hypothesis that $\prec$ be [[well-founded relation|well-founded]], then the statement is a theorem ([[Mostowski's collapsing lemma]]), while the converse is the axiom of foundation.

If you adopt the axiom of anti-foundation (with the strongest notion of extensional relation) instead of foundation, then the universe of [[pure sets]] becomes the [[corecursion|corecursively]] defined ill-founded sets instead of the [[recursion|recursively]] defined well-founded sets.

\hypertarget{structural_meaning}{}\subsection*{{Structural meaning}}\label{structural_meaning}

Since the axiom of foundation is about pure sets, there seems little point to it in a [[structural set theory]]. However, it does have a structural consequence: every set $S$ is the underlying set of elements of a well-founded model for a pure set (in any of the ways described at [[pure set]]). If one assumes the [[axiom of choice]], however, then this statement follows from the [[well-ordering theorem]], since in that case $S$ is the underlying set of a model for a von Neumann [[ordinal number]]. But the axiom of foundation has no stronger structural consequence, since this statement already suffices to ensure that a model of structural set theory can be reconstructed from the material set theory consisting of its well-founded pure sets.

That this statement is the correct structural version of antifoundation may be justified by appeal to the [[material-structural adjunction]].

category: foundational axiom

[[!redirects axiom of foundation]]

[[!redirects axiom of anti-foundation]] [[!redirects axiom of antifoundation]] [[!redirects anti-foundation]] [[!redirects antifoundation]]



\end{document}