Antisymmetric Quotient of Preordered Set is Ordered Set

Theorem
Let $\left({S, \precsim}\right)$ be a preordered set.

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

Let $\left({S / {\sim}, \preceq}\right)$ be the antisymmetric quotient of $\left({S, \precsim}\right)$.

Then:


 * $\left({S / {\sim}, \preceq}\right)$ is an ordered set.


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

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


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

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

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

To show that $\preceq$ is an Definition:ordering, we must show that it is transitive, reflexive, and antisymmetric.

Transitive
Let $P, Q, R \in S / {\sim}$.

Suppose that $P \preceq Q$ and $Q \preceq R$.

Then for some $p \in P$, $q_1, q2 \in Q$, and $r \in R$:


 * $p \precsim q_1$ and $q_2 \precsim r$.

By the definition of quotient set, $q_1 \sim q_2$.

By the definition of $\sim$:


 * $q_1 \precsim q_2$

Since $p \precsim q_1$, $q_1 \precsim q_2$, $q_2 \precsim r$, and $\precsim$ is transitive, Transitive Chaining shows that:


 * $p \precsim r$

Thus by the definition of $\preceq$:


 * $P \preceq R$.

Since this holds for all such $P$, $Q$, and $R$, $\preceq$ is transitive.

Reflexive
Let $P \in S / {\sim}$.

By the definition of quotient set, $P$ is non-empty.

Thus there exists a $p \in P$.

Since $\precsim$ is a preordering, it is reflexive, so $p \precsim p$.

By the definition of $\preceq$, $P \preceq P$.

As this holds for all $P \in S / {\sim}$, $\preceq$ is reflexive.

Antisymmetric
Let $P, Q \in S / {\sim}$ such that:


 * $P \preceq Q$
 * $Q \preceq P$

By the definition of $\preceq$, there are elements $p_1, p_2 \in P$ and $q_1, q_2 \in Q$ such that:


 * $(1)\quad p_1 \precsim q_1$
 * $(2)\quad q_2 \precsim p_2$

Let $p \in P$.

Then by the definition of quotient set:


 * $p \sim p_1$
 * $p_2 \sim p$

By the definition of $\sim$:


 * $p \precsim p_1$
 * $p_2 \precsim p$

Thus by $(1)$ and $(2)$ and the fact that $\precsim$ is transitive:


 * $p \precsim q_1$
 * $q_2 \precsim p$

By the definition of quotient set:


 * $q_1 \sim q_2$

Thus by the definition of $\sim$:


 * $q_1 \precsim q_2$

Since $\precsim$ is transitive:


 * $p \precsim q_2$

We already know that:


 * $q_2 \precsim p$

Thus $p \sim q_2$.

By the definition of quotient set, $p \in Q$.

The same argument shows that each element of $Q$ is also in $P$.

Thus by the Axiom of Extension, $P = Q$.

As this holds for all such $P, Q \in S / {\sim}$, $\preceq$ is antisymmetric.

Relation between Sets implies all their Elements are Related
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:


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