Relation on Empty Set is Equivalence

Theorem
Let $$S = \varnothing$$, that is, the empty set.

Let $$\mathcal R \subseteq S \times S$$ be a relation on $$S$$.

Then $$\mathcal R$$ is the null relation and is an equivalence relation.

Proof
As $$S = \varnothing$$, we have from Cartesian Product Null that $$S \times S = \varnothing$$.

Then it follows that $$\mathcal R \subseteq S \times S = \varnothing$$.

Reflexivity
From the definition:
 * $$\mathcal R = \varnothing \implies \forall x \in S: \left({x, x}\right) \notin \mathcal R$$

But as $$\neg \, \exists x \in S$$ it follows vacuously that $$\mathcal R$$ is reflexive.

Symmetry
It follows vacuously that:
 * $$\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R$$

and so $$\mathcal R$$ is symmetric.

Transitivity
It follows vacuously that:
 * $$\left({x, y}\right), \left({y, z}\right) \in \mathcal R \implies \left({x, z}\right) \in \mathcal R$$

and so $$\mathcal R$$ is transitive.

It follows from the definition that $$\mathcal R$$ is an equivalence relation.