Restriction of Antitransitive Relation is Antitransitive

Theorem

Let $S$ be a set.

Let $\RR \subseteq S \times S$ be an antitransitive relation on $S$.

Let $T \subseteq S$ be a subset of $S$.

Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.

Then $\RR {\restriction_T}$ is an antitransitive relation on $T$.

Proof

Suppose $\RR$ is antitransitive on $S$.

Then by definition:

$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \notin \RR$

So:

\(\ds \set {\tuple {x, y}, \tuple {y, z} }\)

\(\subseteq\)

\(\ds \RR {\restriction_T}\)

\(\ds \leadsto \ \ \)

\(\ds \set {\tuple {x, y}, \tuple {y, z} }\)

\(\subseteq\)

\(\ds \paren {T \times T} \cap \RR\)

Definition of Restriction of Relation

\(\ds \leadsto \ \ \)

\(\ds \tuple {x, z}\)

\(\notin\)

\(\ds \paren {T \times T} \cap \RR\)

as $\RR$ is antitransitive on $S$

\(\ds \leadsto \ \ \)

\(\ds \tuple {x, z}\)

\(\in\)

\(\ds \RR {\restriction_T}\)

Definition of Restriction of Relation

Therefore, if $x, y, z \in T$, it follows that:

$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR {\restriction_T} \implies \tuple {x, z} \notin \RR {\restriction_T}$

and so by definition $\RR {\restriction_T}$ is an antitransitive relation on $T$.

$\blacksquare$

Also see

• Properties of Restriction of Relation‎ for other similar properties of the restriction of a relation.