The identity relation ${\mathrm{\Delta }}_A=\{(a,b)\in A\times A:a=b\}$ 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

${\mathrm{\Delta }}_A:=\{(a,b)\in A\times A:a=b\},{\in }_A:=\{(a,b)\in A\times A:a\in b\},$

sets by The Axiom Schema of Separation: for each formula $\phi$, $\mathrm{\forall }\bar{p}\mathrm{\forall }x\mathrm{\exists }y\mathrm{\forall }z(z\in y\leftrightarrow (z\in x\wedge \phi (z,\bar{p})))$ applied inside $A\times A$ (The Cartesian product $A\times B:=\{z\in \mathcal{P}(\mathcal{P}(A\cup B)):\mathrm{\exists }a\in A\mathrm{\exists }b\in Bz=(a,b)\}$), and relations on $A$ in the sense of Relation, $\mathrm{dom}R$, $\mathrm{ran}R$, $\mathrm{fld}R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$". Thus $(a,b)\in {\mathrm{\Delta }}_A$ 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 $(a,b):=\{\{a\},\{a,b\}\}$).

• ${\mathrm{\Delta }}_A$ carries the set $A$ with it. ${\mathrm{\Delta }}_A$ and ${\mathrm{\Delta }}_B$ are different relations when $A\mathrm{\ne }B$, because $\mathrm{dom}{\mathrm{\Delta }}_A=\mathrm{ran}{\mathrm{\Delta }}_A=A$. That is what makes ${\mathrm{\Delta }}_A$ 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, $x\mathrm{\notin }x$ 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.