Restriction of Antitransitive Relation is Antitransitive

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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 $T$.

$\blacksquare$


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