Restriction of Antireflexive Relation is Antireflexive

Theorem

Let $S$ be a set.

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


Proof

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

Then by definition of antireflexive:

$\forall x \in S: \tuple {x, x} \notin \RR$

We are given $T$ is a subset of $S$, so:

$\forall x \in T: \tuple {x, x} \notin \RR$

We are given $\RR {\restriction_T} \subseteq T \times T$, so:

$\forall x \in T: \tuple {x, x} \notin \RR \restriction_T$

Hence by definition, $\RR {\restriction_T}$ is antireflexive on $T$.

$\blacksquare$


Also see

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