Equivalence of Definitions of Total Ordering

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Theorem

The following definitions of the concept of Total Ordering are equivalent:

Definition 1

$\RR$ is a total ordering on $S$ if and only if:

$(1): \quad \RR$ is an ordering on $S$
$(2): \quad \RR$ is connected

That is, $\RR$ is an ordering with no non-comparable pairs:

$\forall x, y \in S: x \mathop \RR y \lor y \mathop \RR x$

Definition 2

$\RR$ is a total ordering on $S$ if and only if:

\(\text {(1)}: \quad\) \(\ds \RR \circ \RR\) \(\subseteq\) \(\ds \RR\)
\(\text {(2)}: \quad\) \(\ds \RR \cap \RR^{-1}\) \(\subseteq\) \(\ds \Delta_S\)
\(\text {(3)}: \quad\) \(\ds \RR \cup \RR^{-1}\) \(=\) \(\ds S \times S\)


Proof

Definition 1 implies Definition 2

Let $\RR$ be an ordering which is also connected.

By definition 2 of ordering:

$\RR \circ \RR = \RR$
$\RR \cap \RR^{-1} = \Delta_S$

and hence by definition of set equality:

$(1): \quad \RR \circ \RR \subseteq \RR$
$(2): \quad \RR \cap \RR^{-1} \subseteq \Delta_S$

By Relation is Connected and Reflexive iff Total, $\RR$ is a total relation.

Thus by Relation is Total iff Union with Inverse is Trivial Relation:

$(3): \quad \RR \cup \RR^{-1} = S \times S$


Hence $\RR$ is a total ordering by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $\RR$ be a relation which fulfils the conditions:

$(1): \quad \RR \circ \RR \subseteq \RR$
$(2): \quad \RR \cap \RR^{-1} \subseteq \Delta_S$
$(3): \quad \RR \cup \RR^{-1} = S \times S$

By definition of transitive relation, we have from $(1)$ that $\RR$ is transitive.

From Relation is Antisymmetric iff Intersection with Inverse is Coreflexive, it follows from $(2)$ that $\RR$ is an antisymmetric relation.

By Relation is Total iff Union with Inverse is Trivial Relation it follows from $(3)$ that $\RR$ is a total relation.

From Relation is Connected and Reflexive iff Total, $\RR$ is both connected and reflexive.


Hence we have that $\RR$ is:

an ordering by definition 1

and hence:

a total ordering by definition 1.

$\blacksquare$


Sources