wellfoundedness and axiom of foundation
Recall that a relation^{} $R$ on a class $C$ is wellfounded if

1.
For any $x\in C$, the collection^{} $\{y\in C\mid yRx\}$ is a set, and

2.
for any nonempty $B\subseteq C$, there is an element $z\in B$ such that if $yRz$, then $y\notin B$.
$z$ is called an $R$minimal element of $B$. It is clear that the membership relation $\in $ in the class of all sets satisfies the first condition above.
Theorem 1.
Given ZF, $\mathrm{\in}$ is a wellfounded relation iff the Axiom of Foundation^{} (AF) is true.
We will prove this using one of the equivalent^{} versions of AF: for every nonempty set $A$, there is an $x\in A$ such that $x\cap A=\mathrm{\varnothing}$.
Proof.
Suppose $\in $ is wellfounded and $A$ a nonempty set. We want to find $x\in A$ such that $x\cap A=\mathrm{\varnothing}$. Since $\in $ is wellfounded, there is a $\in $minimal set $x$ such that $x\in A$. Since no set $y$ such that $y\in x$ and $y\in A$ (otherwise $x$ would not be $\in $minimal^{}), we have that $x\cap A=\mathrm{\varnothing}$.
Conversely, suppose that AF is true. Let $A$ be any nonempty set. We want to find a $\in $minimal element in $A$. Let $x\in A$ such that $x\cap A=\mathrm{\varnothing}$. Then $x$ is $\in $minimal in $A$, for otherwise there is $y\in A$ such that $y\in x$, which implies $y\in x\cap A=\mathrm{\varnothing}$, a contradiction^{}. ∎
Title  wellfoundedness and axiom of foundation 

Canonical name  WellfoundednessAndAxiomOfFoundation 
Date of creation  20130322 17:25:34 
Last modified on  20130322 17:25:34 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 03E30 