# Symmetric and Transitive Relation is not necessarily Reflexive

## Theorem

Let $S$ be a set.

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

Let $\alpha$ be both symmetric and transitive.

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

## Proof 1

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

Let:

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

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

However, we have:

- $\neg c \mathrel \alpha c$

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

$\blacksquare$

## Proof 2

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

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

- $\forall x, y \in S: x \mathrel \alpha y \iff x = y = 0$

\(\ds x\) | \(\alpha\) | \(\ds y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y = 0\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds x = 0\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(\alpha\) | \(\ds x\) |

Thus $\alpha$ is symmetric.

Now let $x \mathrel \alpha y$ and $y \mathrel \alpha z$

Then:

- $x = y = 0, y = z = 0$

and so

- $x \mathrel \alpha z$

Now let $x = \Z$ such that $x \ne 0$.

Then it is not the case that:

- $x \mathrel \alpha x$

and so $\alpha$ is not reflexive.

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

$\blacksquare$

## Proof 3

Let $S = \set {1, 2}$ be a set.

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

- $\forall x, y \in S: x \mathrel \RR y \iff x = y = 2$

\(\ds x\) | \(\RR\) | \(\ds y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y = 2\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds x = 2\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(\RR\) | \(\ds x\) |

Thus $\alpha$ is symmetric.

Now let $x \mathrel \RR y$ and $y \mathrel \RR z$

Then:

- $x = y = 2, y = z = 2$

and so:

- $x \mathrel \RR z$

Now let $x = 1$.

Then it is not the case that:

- $x \mathrel \RR x$

and so $\RR$ is not reflexive.

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

$\blacksquare$

## Examples

### Subset of Cartesian Plane

The subset of the Cartesian plane defined as:

- $\RR := \set {\tuple {x, y} \in \R^2: -1 \le x \le 1, -1 \le y \le 1}$

determines a relation on $\R^2$ which is symmetric and transitive but not reflexive.

## Sources

- 1979: John E. Hopcroft and Jeffrey D. Ullman:
*Introduction to Automata Theory, Languages, and Computation*... (previous) ... (next): Chapter $1$: Preliminaries: Exercises: $1.9$