Minimally Inductive Set is Ordinal
Theorem
Let $\omega$ denote the minimally inductive set.
Then $\omega$ is an ordinal.
Proof 1
The minimally inductive set is a set of ordinals by definition.
From the corollary of ordinals are well-ordered, it is seen that $\left({\omega, \Epsilon \! \restriction_\omega}\right)$ is a strictly well-ordered set.
It is to be shown by induction on minimally inductive set that $\forall n \in \omega: \omega_n = n$
Basis for the Induction
\(\ds \omega_\O\) | \(=\) | \(\ds \set {x \in \omega: x \in \O}\) | Definition of Initial Segment | |||||||||||
\(\ds \) | \(=\) | \(\ds \O\) | Definition of Empty Set |
Induction Hypothesis
Suppose that $\omega_n = n$ for some $n \in \omega$.
Induction Step
\(\ds \omega_{n^+}\) | \(=\) | \(\ds \set {x \in \omega: x \in n^+}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in \omega: x \in n \lor x \in \set n}\) | Definition of Successor Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in \omega: x \in n \lor x = n}\) | Definition of Singleton | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in \omega: x \in n} \cup \set {x \in \omega: x = n}\) | Definition of Set Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \omega_n \cup \set n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \cup \set n\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds n^+\) | Definition of Successor Set |
And so $\omega$ is an ordinal.
$\blacksquare$
Proof 2
Let $K_I$ denote the set of all nonlimit ordinals.
Let $\On$ denote the set of all ordinals.
Let $a \in \omega$.
It follows that $a^+ \subseteq K_I$, so $a \in K_I$.
Thus:
- $\omega \subseteq K_I \subseteq \On$
We now must prove that $\omega$ is a transitive set, at which point it will satisfy the Alternative Definition of Ordinal.
Let $x \in y$ and $y \in \omega$.
Then:
- $y \in \On \land y^+ \subseteq K_I$
Because $y$ is an ordinal, it is transitive.
Therefore:
- $x \subseteq y$
and:
- $x^+ \subseteq y^+ \subseteq K_I$
Therefore, $x^+ \subseteq K_I$.
Applying the definition of minimally inductive set:
- $x \in \omega$
so $\omega$ is transitive.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 7.32$