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

By Ordinal Membership is Trichotomy:

- $x^+ \in y^+$

$\blacksquare$

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

By Ordinal Membership is Trichotomy:

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

$\blacksquare$