# Symmetric and Transitive Relation is not necessarily Reflexive/Proof 2

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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 symmetric and transitive.

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

## Proof

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.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $3$