Burali-Forti Paradox

Theorem
Let $\operatorname{On}$ denote the class of all ordinals.

Then $\operatorname{On}$ is a proper class:
 * $\operatorname{On} \notin U$

This is one of many paradoxes of naive set theory, which allows all classes to be sets.

Proof
$\in$ is an ordering relation on the ordinal classes.

Thus an ordinal is not an element of itself.

That is:
 * $(1): \quad \forall A \in \operatorname{On}: A \notin A$

However, the class of all ordinal numbers is an ordinal itself.

Since for ordering relations on the ordinals, the membership relation is equivalent to the subset relation in all instances (see the definition of Ordinals):


 * $(2): \quad \forall x \in \operatorname{On}: x \subset \operatorname{On}$

The segment of the class of ordinals is:


 * $(3): \quad \left\{{x \in \operatorname{On}: x \subset \operatorname{On}}\right\}$

which, by $(2)$, is equal to the $\operatorname{On}$.

Therefore $\operatorname{On}$ is an Ordinal.

Suppose that $\operatorname{On} \in U$.

Then it follows that:


 * $\operatorname{On} \in \operatorname{On}$

But then, by $(1)$, it would follow that:


 * $\operatorname{On} \notin \operatorname{On}$

From this contradiction it follows that:
 * $\operatorname{On} \notin U$

Also see

 * The Ordinal Class, $\operatorname{On}$
 * Russell's Paradox, another paradox in naive set theory.