# Set is Ordinal iff Every Transitive Proper Subset is Element of it

## Theorem

Let $x$ be a set.

Then:

- $x$ is an ordinal

- every transitive proper subset of $x$ is an element of $x$.

## Proof

### Necessary Condition

Let $\alpha$ be an arbitrary ordinal.

Let $y$ be a proper subset of $\alpha$ such that $y$ is transitive.

From Transitive Set of Ordinals is Ordinal it follows that $y$ is an ordinal.

By definition of usual ordering of ordinals:

- $y < \alpha$

Hence from Strict Ordering of Ordinals is Equivalent to Membership Relation:

- $y \in \alpha$

As $\alpha$ is arbitrary, it follows that every transitive proper subset of $\alpha$ is an element of $\alpha$.

$\Box$

### Sufficient Condition

Let $x$ be a set such that every transitive proper subset of $x$ is an element of $x$.

Let $\alpha$ be the smallest ordinal which is not in $x$.

This is always possible by Well-Ordering of Class of All Ordinals under Subset Relation.

Thus every ordinal which is smaller than $\alpha$ is an element of $x$.

Hence:

- $\alpha \subseteq x$

Suppose $\alpha$ is a proper subset of $x$.

From Ordinal is Transitive, $\alpha$ is then a transitive proper subset of $x$.

Hence by hypothesis $\alpha \in x$

But we have that $\alpha \notin x$.

Hence it must be the case that $\alpha = x$.

Thus $x$ is an ordinal.

The result follows.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Theorem $2.3$