Trivial Ordering is Universally Compatible
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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.
$\blacksquare$