Successor Set of Ordinal is Ordinal/Proof 2
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
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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 $A$.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 7.24$