The image $R[A]$ and the preimage $R^{-1}[B]$ of a set under a relation

Let $R$ be a relation (Relation, $\mathrm{dom}R$, $\mathrm{ran}R$, $\mathrm{fld}R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$") and let $A$ and $B$ be sets. The image of $A$ under $R$ and the preimage of $B$ under $R$ are

$R[A]:=\{b\in \mathrm{ran}R:\mathrm{\exists }a(a\in A\wedge (a,b)\in R)\},R^{-1}[B]:=\{a\in \mathrm{dom}R:\mathrm{\exists }b(b\in B\wedge (a,b)\in R)\},$

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 $\mathrm{ran}R$ and $\mathrm{dom}R$ respectively. Thus $b\in R[A]$ holds if and only if $(a,b)\in R$ for some $a\in A$, and $a\in R^{-1}[B]$ holds if and only if $(a,b)\in R$ for some $b\in B$ (The Kuratowski ordered pair $(a,b):=\{\{a\},\{a,b\}\}$).

The notation is consistent: applying the first clause to the inverse relation $R^{-1}$ of The inverse relation $R^{-1}$, the composite $S\circ R$, and the restriction $R\upharpoonright A$ gives $R^{-1}[B]=\{a\in \mathrm{ran}(R^{-1}):\mathrm{\exists }b(b\in B\wedge (b,a)\in R^{-1})\}$, which has exactly the elements just described, since $(b,a)\in R^{-1}$ says $(a,b)\in R$ and $\mathrm{ran}(R^{-1})=\mathrm{dom}R$.

• The bracket is not an application. $R[A]$ is the set of things related to something in $A$; it is written with square brackets precisely so that it is never confused with a value $R(a)$, which for a general relation does not exist. Even for a function, $f[S]$ and $f(S)$ mean different things, and only the bracket form is used here.

• Image and range. Taking $A:=\mathrm{dom}R$ gives $R[\mathrm{dom}R]=\mathrm{ran}R$, and taking $A:=\mathrm{\varnothing }$ gives $R[\mathrm{\varnothing }]=\mathrm{\varnothing }$, since no $a$ lies in $\mathrm{\varnothing }$.