Symmetric Transitive and Serial Relation is Reflexive

Theorem

Let $\RR$ be a relation which is:

symmetric
transitive
serial.


Then $\RR$ is reflexive.


Thus such a relation is an equivalence.


Proof

Let $S$ be a set on which $\RR$ is a relation which is symmetric, transitive and serial.


As $\RR$ is symmetric:

$x \mathrel \RR y \implies y \mathrel \RR x$

As $\RR$ is transitive:

$x \mathrel \RR y \land y \mathrel \RR x \implies x \mathrel \RR x$

As $\RR$ is serial:

$\forall x \in S: \exists y \in S: x \mathrel \RR y$


Let $x \in S$.

Then

\(\ds \exists y \in S: \, \) \(\ds \tuple {x, y}\) \(\in\) \(\ds \RR\) as $\RR$ is serial
\(\ds \leadsto \ \ \) \(\ds \tuple {y, x}\) \(\in\) \(\ds \RR\) as $\RR$ is symmetric
\(\ds \leadsto \ \ \) \(\ds \tuple {x, x}\) \(\in\) \(\ds \RR\) as $\RR$ is transitive

Thus:

$\forall x: x \mathrel \RR x$

and by definition $\RR$ is reflexive.


It follows by definition that such a relation is an equivalence relation.

$\blacksquare$


Sources

  • 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $2 \ \text{(c)}$
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Exercise $10.5$