# Order-Extension Principle

## Theorem

Let $S$ be a set.

Let $\preceq$ be an ordering on $S$.

Then there exists a total ordering $\le$ on $S$ such that:

$\forall a, b \in S: \paren {a \preceq b \implies a \le b}$

### Strict Orderings

Let $S$ be a set.

Let $\prec$ be a strict ordering on $S$.

Then there exists a strict total ordering $<$ on $S$ such that:

$\forall a, b \in S: a \prec b \implies a < b$

## Proof 1

Let $\preceq$ be an ordering on the set $S$.

If $\preceq$ is a total ordering, the result is complete.

Suppose, then, that $\preceq$ is not a total ordering.

Let $\TT$ be the set of orderings on $S$ that extend $\preceq$, ordered by inclusion.

Let $C$ be a chain in $T$.

By Union of Chain of Orderings is Ordering, $\bigcup C$ is an ordering.

Thus every chain in $\TT$ has an upper bound in $\TT$.

By Zorn's Lemma, $\TT$ has a maximal element, $\RR$.

$\RR$ is seen to be the total ordering whose existence is to be demonstrated, as follows:

Suppose that:

$\exists a, b \in S: \tuple {a, b} \notin \RR \land \tuple {b, a} \notin \RR$

Let $\RR'$ be the relation defined as:

$\RR' := \RR \cup \set {\tuple {a, b} }$

Let $\RR'^-$ be the transitive closure of $\RR'$.

Then by Ordering can be Expanded to compare Additional Pair, $\RR'^-$ is an ordering.

But $\RR'^- \supsetneq \RR$, contradicting the maximality of $\RR$.

Thus, $\RR$ is a total ordering.

$\blacksquare$

## Proof 2

Let $\prec$ be the reflexive reduction of $\preceq$.

By Reflexive Reduction of Ordering is Strict Ordering, $\prec$ is a strict ordering.

By the strict form of the Order-Extension Principle, there exists a strict total ordering $<$ on $S$ such that:

$\forall a, b \in S: \paren {a \prec b \implies a < b}$

Let $\le$ be the reflexive closure of $<$.

Let $a, b \in S$.

If $a \preceq b$, then by Law of Excluded Middle either $a \prec b$ or $a = b$.

If $a = b$, then by the definition of reflexive closure, $a \le b$.

If $a \prec b$, then by the choice of $<$, $a < b$ so $a \le b$.

Thus for all $a, b \in S$, $a \preceq b \implies a < b$.

$\blacksquare$

## Remarks

As shown in Proof 2, the order-extension principle is weaker than the Boolean Prime Ideal Theorem (BPI).

It is known that it is in fact strictly weaker than BPI.

However, it cannot be proved in Zermel-Fraenkel set theory without the Axiom of Choice. In fact it is known to be strictly stronger than the Ordering Principle.