Symmetric and Transitive Relation is not necessarily Reflexive/Proof 2
Jump to navigation
Jump to search
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 $3$