# Restriction of Antitransitive Relation is Antitransitive

## Theorem

Let $S$ be a set.

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

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

Let $\mathcal R \restriction_T \ \subseteq T \times T$ be the restriction of $\mathcal R$ to $T$.

Then $\mathcal R \restriction_T$ is an antitransitive relation on $T$.

## Proof

Suppose $\mathcal R$ is antitransitive on $S$.

Then by definition:

$\left\{ {\left({x, y}\right), \left({y, z}\right)}\right\} \subseteq \mathcal R \implies \left({x, z}\right) \notin \mathcal R$

So:

 $\displaystyle \left\{ {\left({x, y}\right), \left({y, z}\right)}\right\}$ $\subseteq$ $\displaystyle \mathcal R \restriction_T$ $\displaystyle \implies \ \$ $\displaystyle \left\{ {\left({x, y}\right), \left({y, z}\right)}\right\}$ $\subseteq$ $\displaystyle \left({T \times T}\right) \cap \mathcal R$ by definition of restriction of relation $\displaystyle \implies \ \$ $\displaystyle \left({x, z}\right)$ $\notin$ $\displaystyle \left({T \times T}\right) \cap \mathcal R$ as $\mathcal R$ is antitransitive on $S$ $\displaystyle \implies \ \$ $\displaystyle \left({x, z}\right)$ $\in$ $\displaystyle \mathcal R \restriction_T$ by definition of restriction of relation

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

$\left\{ {\left({x, y}\right), \left({y, z}\right)}\right\} \subseteq \mathcal R \restriction_T \implies \left({x, z}\right) \notin \mathcal R \restriction_T$

and so by definition $\mathcal R \restriction_T$ is an antitransitive relation on $T$.

$\blacksquare$