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 $T$.

$\blacksquare$