Relation on Empty Set is Equivalence

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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 is Empty iff Factor is Empty 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.

$\Box$


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.

$\Box$


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.

$\Box$

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

$\blacksquare$