To show that $\precsim$ is a preordering we must show that it is reflexive and transitive. === Reflexive === Let $p \in S$. Then $\map \phi p = \map \phi p$. Since $\preceq$ is an ordering it is reflexive. Thus $\map \phi p \preceq \map \phi p$. By the definition of $\precsim$, $p \precsim p$. As this holds for all $p \in S$, $\precsim$ is reflexive. {{qed|lemma}} === Transitive === Let $p, q, r \in S$. Suppose that $p \precsim q$ and $q \precsim r$. By the definition of $\precsim$: :$\map \phi p \preceq \map \phi q$ :$\map \phi q \preceq \map \phi r$ Since $\preceq$ is an ordering, it is transitive, so: :$\map \phi p \preceq \map \phi r$ Thus by the definition of $\precsim$, $p \precsim r$. As this holds for all such $p$, $q$, and $r$, $\precsim$ is transitive. {{qed|lemma}} === $\precsim$ induces the $\preceq$ ordering on $\PP$ === Let $P, Q \in P$. First suppose that $P \preceq Q$. Let $p \in P$ and let $q \in Q$. By the definition of quotient mapping: :$\map \phi p = P$ and $\map \phi q = Q$. Thus $\map \phi p \preceq \map \phi q$. So by the definition of $\precsim$: :$p \precsim q$ Suppose instead that for some $p \in P$ and $q \in Q$, $p \precsim q$. Then by the definition of $\precsim$: :$\map \phi p \preceq \map \phi q$ By the definition of quotient mapping: :$\map \phi p = P$ and $\map \phi q = Q$. Thus $P \preceq Q$. {{qed|lemma}} === $\struct {\PP, \preceq}$ is the Antisymmetric Quotient of $\struct {S, \precsim}$ === Let $\sim$ be the equivalence relation on $S$ induced by $\precsim$. First we show that $\PP = S / {\sim}$. As both $\PP$ and $S/ {\sim}$ are partitions of $S$, we need only show that for $p, q \in S$: :$\map \phi p = \map \phi q \iff p \sim q$ First suppose that $p \sim q$. By the definition of $\sim$: :$p \precsim q$ and $q \precsim p$. Then by the definition of $\precsim$: :$\map \phi p \preceq \map \phi q$ :$\map \phi q \preceq \map \phi p$ Since $\preceq$ is an ordering, and hence antisymmetric: :$\map \phi p = \map \phi q$ Suppose instead that $\map \phi p = \map \phi q$. Since $\preceq$ is reflexive: :$\map \phi p \preceq \map \phi q$ :$\map \phi q \preceq \map \phi p$ By the definition of $\precsim$: :$p \precsim q$ and $q \precsim p$. By the definition of $\sim$: :$p \sim q$ Now we must show that for $P, Q \in \PP$: :$P \preceq Q \iff \exists p \in P: \exists q \in Q: p \precsim q$ Suppose that $P \preceq Q$. Let $p \in P$ and $q \in Q$. Then $\map \phi p = P$ and $\map \phi q = Q$, so $p \precsim q$ by the definition of $\precsim$. Suppose instead that $\exists p \in P: \exists q \in Q: p \precsim q$. Then by the definition of $\precsim$, $P \preceq Q$. {{qed|lemma}} === $\precsim$ is Unique === Let $\precsim'$ be a preordering such that the antisymmetric quotient of $\struct {S, \precsim'}$ is $\struct {P, \preceq}$. Let $p, q \in S$. First suppose that $p \precsim' q$. Then $\map \phi p \preceq \map \phi q$ by the definition of antisymmetric quotient. Thus $p \precsim q$ by the definition of $\precsim$. Suppose instead that $p \precsim q$. Then $\map \phi p \preceq \map \phi q$ by the definition of $\precsim$. Thus $p \precsim' q$ by the definition of antisymmetric quotient. {{qed}} Category:Order Theory Category:Preorder Theory sxc05rmv5q06ns1feh0ltljtnph65pw