Reflexive and Transitive Relation is not necessarily Symmetric

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Theorem

Let $S$ be a set.

Let $\alpha \subseteq S \times S$ be a relation on $S$.

Let $\alpha$ be both reflexive and transitive.


Then it is not necessarily the case that $\alpha$ is also symmetric.


Proof 1

Proof by Counterexample:

Let $S = \set {a, b, c}$.

Let:

$\alpha = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, c}, \tuple {a, c} }$

By inspection it is seen that $\alpha$ is both reflexive and transitive.

However, we have:

$a \mathrel \alpha b$

but it is not the case that $b \mathrel \alpha a$.

Hence $\alpha$ is both reflexive and transitive but not symmetric.

$\blacksquare$


Proof 2

Proof by Counterexample:

Let $S = \Z$ be the set of integers.


Let $\alpha$ be the relation on $S$ defined as:

$\forall x, y \in S: x \mathrel \alpha y \iff x \le y$


It is seen that:

$\forall x \in \Z: x \le x$

and so:

$\forall x \in \Z: x \mathrel \alpha x$

Thus $\alpha$ is reflexive.


Then it is seen that:

$\forall x, y, z \in \Z: x \le y, y \le z \implies x \le z$

Thus $\alpha$ is transitive.


Now let $x = 1$ and $y = 2$.

Then:

$x \le y$ but it is not the case that $y \le x$

and so $\alpha$ is not symmetric.


Hence $\alpha$ is both reflexive and transitive but not symmetric.

$\blacksquare$