# Counting Theorem

## Theorem

Every well-ordered set is order isomorphic to a unique ordinal.

## Proof 1

### Existence

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.

Hence from Condition for Woset to be Isomorphic to Ordinal‎, $S_a$ itself is isomorphic to an ordinal.

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

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

$\Box$

### Uniqueness

Uniqueness follows from Isomorphic Ordinals are Equal.

Hence the result.

$\blacksquare$

## Proof 2

Let $A$ be a properly well-ordered class.

Let $\On$ denote the class of all ordinals.

By the Comparability Theorem, either:

$A$ is order isomorphic to a lower section of $\On$

or:

$\On$ is order isomorphic to a lower section of $A$.

Let $A$ be a set.

As $A$ is a set, every lower section of $A$ is a set.

Aiming for a contradiction, suppose $\On$ is order isomorphic to a lower section $L$ of $A$.

Then $\On$ would then be in one-to-one correspondence with that lower section.

Thus there would be a mapping $\phi: L \to \On$.

By the Axiom of Replacement, $\phi \sqbrk L = \On$ is then a set.

But from Class of All Ordinals is Proper Class, $\On$ is a proper class.

From this contradiction it follows that $\On$ cannot be order isomorphic to a lower section $L$ of $A$.

Hence $A$ is order isomorphic to a lower section of $\On$.

From Lower Section of Class of All Ordinals is Ordinal, that means $A$ is order isomorphic to an ordinal.

From Distinct Ordinals are not Order Isomorphic it follows that $A$ is order isomorphic to exactly one such ordinal.

$\blacksquare$

## Also presented as

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

## Motivation

What we have achieved with the Counting Theorem is that for any properly well-ordered collection $A$, we can assign ordinal numbers as indices of the elements of $A$, treating the latter as a sequence:

the $1$st, the $2$nd, $\ldots$, the $\alpha$th, $\ldots$ elements of $A$

and moreover, we can assign these indices in an order-preserving way.

That is, for all ordinal numbers $\alpha$ and $\beta$, $\alpha < \beta$ if and only if the $\alpha$th element comes before the $\beta$th element.