Successor is Less than Successor/Sufficient Condition/Proof 2
Jump to navigation
Jump to search
![]() | This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. 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 {{MissingLinks}} from the code. |
Theorem
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
- $x \in y$
Proof
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$.