# 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: $x^+ \in y^+$.

$\blacksquare$