Subset is Compatible with Ordinal Successor/Proof 3
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.
By Ordinal Membership is Trichotomy, one of the following must be true:
| \(\ds x^+\) | \(=\) | \(\ds y^+\) | ||||||||||||
| \(\ds y^+\) | \(\in\) | \(\ds x^+\) | ||||||||||||
| \(\ds x^+\) | \(\in\) | \(\ds y^+\) |
We will show that the first two are both false, so that the third must hold.
Two preliminary facts:
| \((1)\) | $:$ | \(\ds x \) | \(\ds \ne \) | \(\ds y \) | $x \in y$ and Ordinal is not Element of Itself | ||||
| \((2)\) | $:$ | \(\ds y \) | \(\ds \notin \) | \(\ds x \) | $x \in y$ and Ordinal Membership is Asymmetric |
By $(1)$ and Equality of Successors:
- $x^+ \ne y^+$
Thus the first of the three possibilities is false.
Aiming for a contradiction, suppose $y^+ \in x^+$.
Then:
| \(\ds y\) | \(\in\) | \(\ds y^+\) | Definition of Successor Set | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds x^+\) | $y^+ \in x^+$ and $x^+$ is an ordinal and therefore transitive. | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds x\) | Definition of Successor Set | ||||||||||
| \(\, \ds \lor \, \) | \(\ds y\) | \(=\) | \(\ds x\) |
But we already know that $y \notin x$ by $(2)$ and $y \ne x$ by $(1)$.
So this is a contradiction, and we conclude that $y^+ \notin x^+$.
Thus we have shown that the second possibility is false.
Thus the third and final one must hold: $x^+ \in y^+$.
$\blacksquare$