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
All ordinals are not elements of themselves, since epsilon is an ordering relation on the ordinal classes.

That is:
 * $(1): \quad \displaystyle \forall A \in \operatorname{On}: A \not\in A$

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), we have that:


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

The segment of the class of ordinals is:


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

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

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

If $\operatorname{On} \in U$, then it follows that:


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

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


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

Therefore, by contradiction, $\operatorname{On} \notin U$.

Also see

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