# Strict Ordering of Ordinals is Equivalent to Membership Relation

## Theorem

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

Let $<$ denote the (strict) usual ordering of $\On$.

Then:

$\forall \alpha, \beta \in \On: \alpha < \beta \iff \alpha \in \beta$

## Proof

### Necessary Condition

Let $\alpha \in \beta$.

Then from Ordinal is Transitive:

$\alpha \subseteq \beta$

But if $\alpha = \beta$ we would have $\alpha \in \alpha$.

This is contrary to Ordinal is not Element of Itself.

Hence we have:

$\alpha \subseteq \beta$

and:

$\alpha \ne \beta$

That is:

$\alpha \subsetneqq \beta$

Hence by definition of the (strict) usual ordering of $\On$:

$\alpha < \beta$

$\Box$

### Sufficient Condition

Let $\alpha < \beta$.

$\alpha^+ \subseteq \beta$

where $\alpha^+$ denotes the successor set of $\alpha$.

By definition of successor set:

$\alpha^+ = \alpha \cup \set \alpha$

and so:

$\alpha \in \alpha^+$

That is:

$\alpha \in \beta$

$\blacksquare$