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

$\blacksquare$