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

\(\displaystyle x \in y\) \(\implies\) \(\displaystyle x \ne y\) No Membership Loops
\(\displaystyle \) \(\implies\) \(\displaystyle x^+ \ne y^+\) Equality of Successors
\(\displaystyle x \in y\) \(\implies\) \(\displaystyle y \notin x\) No Membership Loops
\(\displaystyle \) \(\implies\) \(\displaystyle y \notin x^+\) $x ≠ y$ and Definition:Successor Set
\(\displaystyle y^+ \in x^+\) \(\implies\) \(\displaystyle y \in x^+\) 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 $x^+ = y^+$.

Suppose for the sake of contradiction that $y^+ = x^+$.

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 $y^+ ≠ x^+$.

Suppose for the sake of contradiction that $y^+ \in x^+$.

By the 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:

$x^+ = y^+$
$y^+ \in x^+$
$x^+ \in y^+$

We will show that the first two are both false, so that the third must hold.


Two preliminary facts:

\((1)\)   $:$   $x \ne y$             $x \in y$ and Ordinal is not Element of Itself
\((2)\)   $:$   $y \notin x$             $x \in y$ and Ordinal Membership is Asymmetric


By $(1)$ and Equality of Successors:

$x^+ ≠ y^+$

Thus the first of the three possibilities is false.


Suppose for the sake of contradiction that $y^+ \in x^+$.

Then:

\(\displaystyle \) \(\) \(\displaystyle y \in y^+\) Definition of successor set
\(\displaystyle \) \(\implies\) \(\displaystyle y \in x^+\) $y^+ \in x^+$ and $x^+$ is an ordinal and therefore transitive.
\(\displaystyle \) \(\implies\) \(\displaystyle y \in x \lor y = x\) Definition of successor set

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

$\blacksquare$