Antisymmetric Quotient of Preordered Set is Ordered Set

Theorem
Let $\struct {S, \precsim}$ be a preordered set.

Let $\sim$ be the equivalence relation $S$ induced by $\precsim$.

Let $\struct {S / {\sim}, \preceq}$ be the antisymmetric quotient of $\struct {S, \precsim}$.

Then:


 * $\struct {S / {\sim}, \preceq}$ is an ordered set.


 * $\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$

This second statement means that we could just as well have defined $\preceq$ by letting $P \preceq Q$ :


 * $\forall p \in P: \forall q \in Q: p \precsim q$

Proof
By the definition of equivalence relation, $\sim$ is transitive, reflexive, and symmetric.

By the definition of preordering, $\precsim$ is transitive and reflexive.

From Preordering induces Ordering, $\preceq$ is an ordering.

Let $P, Q \in S / {\sim}$ with $P \preceq Q$.

Then by the definition of $\preceq$, there are $p \in P$ and $q \in Q$ such that $p \precsim q$.

Let $p' \in P$ and $q' \in Q$.

By the definition of quotient set:


 * $p' \sim p$
 * $q \sim q'$

Thus by the definition of $\sim$:


 * $p' \precsim p$
 * $q \precsim q'$

Since $p \precsim q$ and $\precsim$ is transitive:


 * $p' \precsim q'$

We have shown that:


 * $\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$