Restriction of Connected Relation is Connected

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Theorem

Let $S$ be a set.

Let $\mathcal R \subseteq S \times S$ be a connected 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 a connected relation on $T$.


Proof

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

That is:

$\forall a, b \in S: a \ne b \implies \left({a, b}\right) \in \mathcal R \lor \left({b, a}\right) \in \mathcal R$


So:

\(\displaystyle a, b\) \(\in\) \(\displaystyle T\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left({a, b}\right)\) \(\in\) \(\displaystyle T \times T\)
\(\, \displaystyle \land \, \) \(\displaystyle \left({b, a}\right)\) \(\in\) \(\displaystyle T \times T\) by definition of ordered pair and cartesian product
\(\displaystyle \implies \ \ \) \(\displaystyle \left({a, b}\right)\) \(\in\) \(\displaystyle \left({T \times T}\right) \cap \mathcal R\)
\(\, \displaystyle \lor \, \) \(\displaystyle \left({b, a}\right)\) \(\in\) \(\displaystyle \left({T \times T}\right) \cap \mathcal R\) as $\mathcal R$ is connected on $S$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({a, b}\right)\) \(\in\) \(\displaystyle R \restriction_T\)
\(\, \displaystyle \lor \, \) \(\displaystyle \left({b, a}\right)\) \(\in\) \(\displaystyle R \restriction_T\) by definition of restriction of relation


and so $\mathcal R \restriction_T$ is connected on $T$.

$\blacksquare$


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