Ordinal is Member of Class of All Ordinals
Jump to navigation
Jump to search
Theorem
Let $A$ be an ordinal.
Then:
- $A \in \On \lor A = \On$
where $\On$ denotes the class of all ordinals.
Proof
We have by hypothesis that $A$ is an ordinal
From Class of All Ordinals is Ordinal and Ordinal Membership is Trichotomy:
- $A \in \On \lor A = \On \lor \On \in A$
But by the Burali-Forti Paradox $\On$ is a proper class.
Therefore:
- $A \in \On \lor A = \On$
$\blacksquare$
This article, or a section of it, needs explaining. In particular: please say or link to why being a proper class means that $\On \notin A$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $7.14$