Diagonal Relation Equivalence
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Theorem
The diagonal relation $\Delta_S$ on $S$ is always an equivalence in $S$.
Proof
Checking in turn each of the criteria for equivalence:
Reflexive
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \forall x \in S: \left({x, x}\right) \in \Delta_S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Diagonal Relation |
So $\Delta_S$ is reflexive.
$\Box$
Symmetric
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \forall x, y \in S: \left({x, y}\right) \in \Delta_S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle x = y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Diagonal Relation | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle y = x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Equality is Symmetric | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \left({y, x}\right) \in \Delta_S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Diagonal Relation |
So $\Delta_S$ is symmetric.
$\Box$
Transitive
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \forall x, y, z \in S: \left({x, y}\right) \in \Delta_S \land \left({y, z}\right) \in \Delta_S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle x = y \land y = z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Diagonal Relation | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle x = z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Equality is Transitive | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \left({x, z}\right) \in \Delta_S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Diagonal Relation |
So $\Delta_S$ is transitive.
$\blacksquare$
Sources
- Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 7$: Relations
- George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$
- T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 6$: Example $6.7$