# Successor Set of Ordinal is Ordinal

## Theorem

Let $\On$ denote the class of all ordinals.

Let $\alpha \in \On$ be an ordinal.

Then its successor set $\alpha^+ = \alpha \cup \set \alpha$ is also an ordinal.

## Proof 1

We have the result that Class of All Ordinals is Minimally Superinductive over Successor Mapping.

Hence $\On$ is *a fortiori* a superinductive class with respect to the successor mapping.

Hence, by definition of superinductive class:

- $\On$ is closed under the successor mapping.

That is:

- $\forall \alpha \in \On: \alpha^+ \in \On$

$\blacksquare$

## Proof 2

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From Ordinal is Transitive, it follows by Successor Set of Transitive Set is Transitive that $\alpha^+$ is transitive.

We now have to show that $\alpha^+$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_{\alpha^+} }$.

So suppose that a subset $A \subseteq \alpha^+$ is non-empty.

Then:

\(\ds A\) | \(=\) | \(\ds A \cap \paren {\alpha \cup \set \alpha}\) | Intersection with Subset is Subset and Definition of Successor Set | |||||||||||

\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {A \cap \alpha} \cup \paren {A \cap \set \alpha}\) | Intersection Distributes over Union |

We need to show that $A$ has a smallest element.

We first consider the case where $A \cap \alpha$ is empty.

By equation $\paren 1$, it follows that $A \cap \set \alpha$ is non-empty (because $A$ is non-empty).

Therefore:

- $\alpha \in A$

That is:

- $\set \alpha \subseteq A$

By Union with Empty Set and Intersection with Subset is Subset, equation $\paren 1$ implies that $A \subseteq \set \alpha$.

Therefore, $A = \set \alpha$ by the definition of set equality.

So $\alpha$ is the smallest element of $A$.

We now consider the case where $A \cap \alpha$ is non-empty.

- $A \cap \alpha \subseteq \alpha$

By the definition of a well-ordered set, there exists a smallest element $x$ of $A \cap \alpha$.

Let $y \in A$.

If $y \in \alpha$, then $y \in A \cap \alpha$

Therefore, by the definition of the smallest element, either $x \in y$ or $x = y$.

Otherwise, $y = \alpha$, and so $x \in \alpha = y$.

That is, $x$ is the smallest element of $A$.

$\blacksquare$