An ordinal number / ordinal is a transitive set that is strictly well-ordered by $\in$.
The class of all ordinals is denoted by $\Ord$.
For two ordinals $\alpha$ and $\beta$, we define the following total order on $\Ord$:
\[\alpha < \beta \enspace\lrimp\enspace \alpha\in\beta.\]
$0:=\varnothing$ is an ordinal.
Every element of an ordinal is an ordinal.
For any two ordinals $\alpha$ and $\beta$:
- \[\alpha\subseteq\beta \enspace\lor\enspace \beta\subseteq\alpha.\]
- \[\alpha\subset\beta \enspace\rimp\enspace \alpha\in\beta.\]
For any ordinal $\alpha$,
\[\alpha = \{\beta:\beta < \alpha\}.\]
If $C$ is a non-empty class of ordinals, then $\bigcap C$ is an ordinal and
\[\bigcap C \,=\, \inf C \,\in\, C.\]
If $X$ is a non-empty set of ordinals, then $\bigcup X$ is an ordinal and
\[\bigcup X = \sup X.\]
ORD#PROP-BF. Burali-Forti Paradox.
$\Ord$ is a proper class.
Proof 1.By ORD#PROP-U, $\sup\Ord$ would be an ordinal, hence
\[(\sup\Ord)+1 \leq \sup\Ord,\]a contradiction.
Proof 2.By CA#PROP-PC, $\Card$ is a proper class. Since $\Card\subset\Ord$, $\Ord$ is a proper class.