Counting Theorem/Proof 1

Theorem

Let $\struct {S, \preceq}$ be a woset.

From Condition for Woset to be Isomorphic to Ordinal‎, it is enough to show that for every $a \in S$, the initial segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal.

Let:

$E = \set {a \in S: S_a \text{ is not isomorphic to an ordinal} }$

We will show that $E = \O$.

Aiming for a contradiction, suppose that $E \ne \O$.

Let $a$ be the minimal element of $E$.

This is bound to exist by definition of woset.

So, if $x \prec a$, it follows that $S_x$ is isomorphic to an ordinal.

But for $x \prec a$, we have $S_x = \paren {S_a}_x$ from definition of an ordinal.

So every segment of $S_a$ is isomorphic to an ordinal.

This contradicts the supposition that $a \in E$.

Hence $E = \O$ and existence has been proved.

$\Box$

Hence the result.

$\blacksquare$

Some sources present the Counting Theorem as the definition of an ordinal as the order type of a well-ordering.

1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.12$