Diagonal Relation is Equivalence

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Theorem

The diagonal relation $\Delta_S$ on a set $S$ is always an equivalence in $S$.


Proof

Checking in turn each of the criteria for equivalence:


Reflexive

\(\, \displaystyle \forall x \in S: \, \) \(\displaystyle \tuple {x, x}\) \(\in\) \(\displaystyle \Delta_S\) $\quad$ Definition of Diagonal Relation $\quad$

So $\Delta_S$ is reflexive.

$\Box$


Symmetric

\(\, \displaystyle \forall x, y \in S: \, \) \(\displaystyle \tuple {x, y}\) \(\in\) \(\displaystyle \Delta_S\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle y\) $\quad$ Definition of Diagonal Relation $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle x\) $\quad$ Equality is Symmetric $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \tuple {y, x}\) \(\in\) \(\displaystyle \Delta_S\) $\quad$ Definition of Diagonal Relation $\quad$

So $\Delta_S$ is symmetric.

$\Box$


Transitive

\(\, \displaystyle \forall x, y, z \in S: \, \) \(\displaystyle \tuple {x, y}\) \(\in\) \(\displaystyle \Delta_S \land \tuple {y, z} \in \Delta_S\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle y \land y = z\) $\quad$ Definition of Diagonal Relation $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle z\) $\quad$ Equality is Transitive $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \tuple {x, z}\) \(\in\) \(\displaystyle \Delta_S\) $\quad$ Definition of Diagonal Relation $\quad$

So $\Delta_S$ is transitive.

$\blacksquare$


Sources