Asymmetric Relation is Antireflexive

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be a set.

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

Let $\mathcal R$ be asymmetric.


Then $\mathcal R$ is also antireflexive.


Proof

Let $\mathcal R$ be asymmetric.

Then, by definition:

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


Suppose $\left({x, x}\right) \in \mathcal R$. Then:

\(\displaystyle \left({x, x}\right) \in \mathcal R\) \(\implies\) \(\displaystyle \left({x, x}\right) \notin \mathcal R\) Definition of Asymmetric Relation
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x, x}\right) \notin \mathcal R\) \(\) \(\displaystyle \) Proof by Contradiction

Thus $\mathcal R$ is antireflexive.

$\blacksquare$


Sources