A relation $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ is total if, for each $a\in A$, we have $R\webleft (a\webright )\neq \text{Ø}$.
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(b)
We have $\chi _{A}\subset R^{\dagger }\mathbin {\diamond }R$.
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Item 1a$\implies $Item 1b: 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 $R$.
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Item 1b$\implies $Item 1a: Given $a\in A$, since $\chi _{A}\subset R^{\dagger }\mathbin {\diamond }R$, we must have
\[ \left\{ a\right\} \subset \webleft [R^{\dagger }\mathbin {\diamond }R\webright ]\webleft (a\webright ), \]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 $b\in R\webleft (a\webright )$.
10.1.2 Total Relations
Let $A$ and $B$ be sets.
Let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be a relation.
Proof of Proposition 10.1.2.1.2.
Item 1: Characterisations
We claim that Item 1a and Item 1b are indeed equivalent:
This finishes the proof.
