Preimage of Singleton

Theorem

Let $\RR$ be a relation.

Let $\map {\RR^{-1} } t$ denote the preimage of $t$ under $\RR$.

Let $\RR^{-1} \sqbrk {\set t}$ denote the preimage of $\set t$ under $\RR$.

Then:

$\RR^{-1} \sqbrk {\set t} = \map {\RR^{-1} } t$

Proof

\(\ds \map {\RR^{-1} } t\)

\(=\)

\(\ds \set {s: \paren {s, t} \in \RR}\)

Definition of Preimage of Element under Relation

\(\ds \)

\(=\)

\(\ds \set {s: \exists y: \paren {y = t \land \tuple {s, y} \in \RR} }\)

Equality implies Substitution

\(\ds \)

\(=\)

\(\ds \set {s: \exists y \in \set t: \tuple {s, y} \in \RR}\)

Definition of Singleton

\(\ds \)

\(=\)

\(\ds \RR^{-1} \sqbrk {\set t}\)

Definition of Preimage of Subset under Relation

$\blacksquare$

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 6.19$