Domain of Relation is Image of Inverse Relation
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Theorem
Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$.
Then:
- $\Dom \RR = \Img {\RR^{-1} }$
That is, the domain of a relation is the image of its inverse.
Proof
By definition:
\(\ds \Dom \RR\) | \(:=\) | \(\ds \set {s \in S: \exists t \in T: \tuple {s, t} \in \RR}\) | ||||||||||||
\(\ds \Img {\RR^{-1} }\) | \(:=\) | \(\ds \set {s \in S: \exists T \in T: \tuple {t, s} \in \RR^{-1} }\) |
\(\ds x\) | \(\in\) | \(\ds \Dom \RR\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists t \in T: \, \) | \(\ds \tuple {x, t}\) | \(\in\) | \(\ds \RR\) | Definition of Domain of Relation | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists t \in T: \, \) | \(\ds \tuple {t, x}\) | \(\in\) | \(\ds \RR^{-1}\) | Definition of Inverse Relation | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \Img {\RR^{-1} }\) | Definition of Image of Relation |
$\blacksquare$
Also see
Sources
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Relations
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 10$: Inverses and Composites
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.11$: Relations