# Axiom of regularity definition

It is stated that:

In mathematics, the axiom of regularity (also known as the axiom of foundation) is an axiom of Zermelo–Fraenkel set theory that states that every non-empty set $A$ contains an element that is disjoint from $A$.

$$\forall x\left(x\neq \emptyset\implies\exists y\in x\left(y\cap x=\emptyset\right)\right)$$

How can this be correct?

If $A$ contains an element $x$, then $x$ can not be disjoint from $A$, because $x$ belongs both to $A$ and to the set containing only $x$.

What am I misunderstanding?

• Your question makes no sense. Syntactically. – Asaf Karagila Jun 1 '17 at 10:33
• What does it mean that $x$ is "from" $A$, and what do you mean by "belongs"? – Asaf Karagila Jun 1 '17 at 10:48
• Are you concerned about an element being part of two or more sets at the same time? Look at this: $A=\{\{1,2\},\{2,3\},2\}$. The $2$ is element of both sets inside $A$ and of $A$ itself. Are you feeling uncomfortable with this? – M. Winter Jun 1 '17 at 10:50
• I updated my question because I phrased it poorly. – octavian Jun 1 '17 at 10:52
• An element $x$ is not the same as the set $\{x\}$ containing only this element. Is this the confusing point? If $x\in A$, then $\{x\}$ must not be in $A$, so $x$ can be disjoint from $A$. – M. Winter Jun 1 '17 at 10:54

One consequence of this is that $x \cap \{x\} = \emptyset$. This often confuses people, because it seems counter-intuitive.
But if the sets $x$ and $\{x\}$ have a common element, it must be $x$ itself, since $x$ is the only element of $\{x\}$. And this would lead us to conclude (reluctantly) that $x \in x$. Such a set would have "no foundation", since we'd have to "keep opening up" the set $x$ only to find still another one inside it: $x\in x \in x \in x\dots$