Subset is Compatible with Ordinal Successor

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

$\blacksquare$