# Subset is Compatible with Ordinal Successor

## 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 1

 $\ds x \in y$ $\implies$ $\ds x \ne y$ No Membership Loops $\ds$ $\implies$ $\ds x^+ \ne y^+$ Equality of Successors $\ds x \in y$ $\implies$ $\ds y \notin x$ No Membership Loops $\ds$ $\implies$ $\ds y \notin x^+$ $x ≠ y$ and Definition of Successor Set $\ds y^+ \in x^+$ $\implies$ $\ds y \in x^+$ Successor Set of Ordinal is Ordinal, Ordinals are Transitive, and Set is Element of Successor

The last part is a contradiction, so $y^+ \notin x^+$.

$x^+ \in y^+$

$\blacksquare$

## Proof 2

First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.

Let $x \in y$.

We wish to show that $x^+ \in y^+$.

Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$.

Aiming for a contradiction, suppose $y^+ = x^+$.

Then $y \in x$ or $y = x$ by the definition of successor set.

If $y = x$ then $x \in x$, contradicting the fact that Ordinal is not Element of Itself.

If $y \in x$ then since an ordinal is transitive, $y \in y$, again contradicting Ordinal is not Element of Itself.

Thus $y^+ ≠ x^+$.

Aiming for a contradiction, suppose $y^+ \in x^+$.

By definition of successor set, $y^+ \in x$ or $y^+ = x$.

If $y^+ \in x$, then since $y^+$ and $x$ are both ordinals, $y^+ \subsetneqq x$.

Then $y \in x$.

Since $y$ is transitive, $y \in y$, contradicting Ordinal is not Element of Itself.

If instead $y^+ = x$, then $x \in y \in y^+ = x$, so the same contradiction arises because $x$ is transitive.

Thus $y^+ \notin x^+$.

So the only remaining possibility, that $x^+ \in y^+$, must hold.

$\blacksquare$

## Proof 3

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

$\blacksquare$