Let $\mathrm M = \Bbb R; \mathrm R = \{(x,y)\mid x = y\}$
Investigate wheter the relation is reflexive, transitive, symemtric, antisymmetric.
- Reflexivity $\rightarrow (\forall x \in \mathrm A)[(x,x) \in \mathrm R] $
- Symetry $\rightarrow (\forall x, y \in \mathrm A)[(x,y) \in \mathrm R \implies (y,x) \in \mathrm R] $
- Antisymetry $\rightarrow (\forall x, y \in \mathrm A)[(x,y) \in \mathrm R \land (y,x) \in \mathrm R \implies x = y ]$
- Transitive $\rightarrow (\forall x, y, z \in \mathrm A)[(x,y) \in \mathrm R \land(y,z) \in \mathrm R \implies (x,z) \in \mathrm R] $
So (...)
- Reflexivity [
TRUE]: $(x,y)\in \mathrm R \implies x=y \implies (x,x) \in \mathrm R$ - Symmetry[
TRUE]: $(x, y)\in \mathrm R \implies x = y \implies (y, x) \in \mathrm R$ - Antisymmetry [
TRUE]: $(x, y) \in \mathrm R \land (y,x) \in \mathrm R \implies x = y $ - Transitivity[
TRUE]: $(x,y) \in \mathrm R \land (y,z) \in \mathrm R \implies x = y \land y = z \implies x = z \implies (x,z) \in \mathrm R$
But there's no way that a relation be Symetric and Antrisymetric.
What is going on here?
Equality is both symmetric and antisymmetric: despite their names, the two properties are not contradictory. A relation cannot be both symmetric and asymmetric ($\langle x,y\rangle\in R$ implies $\langle y,x\rangle\notin R$), but as you’ve observed, a relation can be both symmetric and antisymmetric. In fact, a relation $R$ on a set $A$ is both symmetric and antisymmetric if and only if it is a subset of the equality relation on $A$; geometrically, it’s a subset of the diagonl in $A\times A$.