Trivial Ordering is Universally Compatible

Theorem

Let $S$ be a set.

Let $\RR$ be the trivial ordering on $S$.

Then $\RR$ is universally compatible.

Proof

To prove that the trivial ordering is in fact an ordering, we need to checking each of the criteria for an ordering:

Reflexivity

$\forall a \in S: a \mathrel \RR a$:

From its definition, we have $\forall a, b \in S: a = b \implies a \mathrel \RR b$.

Thus, as $a = a$, we have $\forall a \in S: a \mathrel \RR a$.

So reflexivity is proved.

Transitivity

$\forall a, b, c \in S: a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$:

From the definition:

$a \mathrel \RR b \iff a = b$
$b \mathrel \RR c \iff b = c$

So as $a = b \land b = c \implies a = c$ from transitivity of equals, we have that $a \mathrel \RR c$ and thus transitivity is proved.

Antisymmetry

$\forall a, b \in S: a \mathrel \RR b \land b \mathrel \RR a \implies a = b$:

From the definition:

$a \mathrel \RR b \iff a = b$
$b \mathrel \RR a \iff b = a$

Antisymmetry follows from symmetry of equals.

The trivial ordering is by definition the same as the diagonal relation, and is therefore universally compatible.

$\blacksquare$