Subsection 6.1.5 (00JJ): Total Relations

Style	Class	Font	Theorem Environments
Style 1	`book`	Alegreya Sans	`tcbthm`
Style 2	`book`	Alegreya Sans	`amsthm`
Style 3	`book`	Arno*	`amsthm`
Style 4	`book`	Computer Modern	`amsthm`

A relation $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ is total if, for each $a\in A$, we have $R\webleft (a\webright )\neq \text{Ø}$.

Let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be a relation.

Characterisations. The following conditions are equivalent:
1. The relation $R$ is total.
2. We have $\chi _{A}\subset R^{\dagger }\mathbin {\diamond }R$.

Item 1: Characterisations

We claim that Item (a) and Item (b) are indeed equivalent:

Item (a)$\implies $Item (b): We have to show that, for each $\webleft (a,a'\webright )\in A$, we have
\[ \chi _{A}\webleft (a,a'\webright )\preceq _{\{ \mathsf{t},\mathsf{f}\} }\webleft [R^{\dagger }\mathbin {\diamond }R\webright ]\webleft (a,a'\webright ), \]

i.e. that if $a=a'$, then there exists some $b\in B$ such that $a\sim _{R}b$ and $b\sim _{R^{\dagger }}a'$ (i.e. $a\sim _{R}b$ again), which follows from the totality of $R$.
Item (b)$\implies $Item (a): Given $a\in A$, since $\chi _{A}\subset R^{\dagger }\mathbin {\diamond }R$, we must have

\[ \webleft\{ a\webright\} \subset \webleft [R^{\dagger }\mathbin {\diamond }R\webright ]\webleft (a\webright ), \]

implying that there must exist some $b\in B$ such that $a\sim _{R}b$ and $b\sim _{R^{\dagger }}a$ (i.e. $a\sim _{R}b$) and thus $R\webleft (a\webright )\neq \text{Ø}$, as $b\in R\webleft (a\webright )$.

This finishes the proof.

The Clowder Project

6.1.5 Total Relations