Minimally Inductive Set is Ordinal/Proof 1
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Theorem
Let $\omega$ denote the minimally inductive set.
Then $\omega$ is an ordinal.
Proof
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.
$\blacksquare$