The identity relation $\Delta_A = \{\,(a,b) \in A \times A : a = b\,\}$ and the membership relation $\in_A\, = \{\,(a,b) \in A \times A : a \in b\,\}$

Let $A$ be a set. The identity relation and the membership relation on $A$ are

$\Delta_A := \{\,(a,b) \in A \times A : a = b\,\}, \qquad \in_A \;:=\; \{\,(a,b) \in A \times A : a \in b\,\},$

sets by The Axiom Schema of Separation: for each formula $\varphi$, $\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))$ applied inside $A \times A$ (The Cartesian product $A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}$), and relations on $A$ in the sense of Relation, $\operatorname{dom} R$, $\operatorname{ran} R$, $\operatorname{fld} R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$". Thus $(a,b) \in \Delta_A$ holds if and only if $a = b$ and $a \in A$, and $(a,b) \in \in_A$ holds if and only if $a \in b$ and both lie in $A$ (The Kuratowski ordered pair $(a,b) := \{\{a\},\{a,b\}\}$).

• $\Delta_A$ carries the set $A$ with it. $\Delta_A$ and $\Delta_B$ are different relations when $A \neq B$, because $\operatorname{dom} \Delta_A = \operatorname{ran} \Delta_A = A$. That is what makes $\Delta_A$ usable as a neutral element for composition on the left and on the right, with the correct set in each position.

• The membership relation is the one Foundation constrains. $\in_A$ is set membership restricted to $A$, so Under Foundation, $x \notin x$ for every set $x$, there are no sets with $x \in y \in x$, and there are no sets with $x \in y \in z \in x$ says that it has no cycle of length one, two or three.