Equivalence of Definitions of Minimally Inductive Set

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition $1$ equals Definition $2$ ===
We will prove that both definitions of $\omega$ specify the same set.
Firstly, let us show that $\omega$ as given in Definition 2 is an inductive set.
It is immediate from the definition of a finite ordinal that $\O$ is one.
Also, if $\beta$ is a finite ordinal, so is $\beta^+$.
Therefore:
:$\omega = \set {\alpha: \alpha \text{ is a finite ordinal} }$
is an inductive set.
Next, apply Definition 1 to this $\omega$, yielding the set:
:$\ds \bigcap \set {S \subseteq \omega: S \text{ is an inductive set} }$
Conversely, suppose that $S \subseteq \omega$ is an inductive set.
Suppose that $S \ne \omega$.
Then their difference $\omega \setminus S$ is non-empty.
By Intersection of Ordinals is Smallest:
:$\ds \alpha := \bigcap \omega \setminus S$
is the smallest element in $\omega \setminus S$.
Then $\alpha \ne \O$, since $\O \in S$ by definition of an inductive set.
So $\alpha = \beta^+$ for some $\beta \in \omega$.
Since $\alpha$ was the smallest element of $\omega \setminus S$, it follows from Ordinal is Less than Successor that:
:$\beta \notin \omega \setminus S$
that, is, $\beta \in S$.
But since $S$ is an inductive set, this means that:
:$\beta^+ = \alpha \in S$
which is a contradiction.
It follows that $S = \omega$.
Therefore, we conclude that:
:$\ds \bigcap \set {S \subseteq \omega: S \text{ is an inductive set} } = \omega$
establishing the equivalence of the definitions.
=== Definition 2 equals Definition 3 ===
From Definition 2 and Definition 3, we see that we are to prove:
:$\alpha$ is a finite ordinal $\alpha^+ \subseteq K_I$
where $\alpha^+$ is the successor ordinal of $\alpha$, and $K_I$ is the class of non-limit ordinals.
Suppose that $\alpha$ is a finite ordinal.
If $\alpha = \O$, then $\alpha^+ = \set \O$.
Since $\O$ is by definition a non-limit ordinal, it follows that:
:$\alpha^+ \subseteq K_I$
Suppose now that $\alpha$ is the smallest finite ordinal such that:
:$\alpha^+ \nsubseteq K_I$
Then since $\alpha \ne \O$, it must be that $\alpha = \beta^+$ for some finite ordinal $\beta$.
Moreover, since $\alpha$ was the smallest such ordinal, we then know that:
:$\alpha = \beta^+ \subseteq K_I$
so that we are led to the conclusion that:
:$\alpha \notin K_I$
However, this contradicts the fact that $\alpha$ is the successor ordinal of $\beta$.
Therefore, $\alpha$ cannot exist, and we conclude that:
:$\alpha^+ \subseteq K_I$ for all finite ordinals $\alpha$.
Conversely, suppose that $\alpha^+ \subseteq K_I$.
Let $\beta \in \alpha^+$ be the smallest element of $\alpha^+$ that is not a finite ordinal, if it exists.
Then by assumption, $\beta \in K_I$, meaning that either $\beta = \O$ or $\beta = \gamma^+$ for some ordinal $\gamma$.
Since $\O$ is a finite ordinal, the former option is impossible.
Thus $\beta = \gamma^+$ for some finite ordinal $\gamma$, by construction of $\beta$ as the smallest element of $\alpha^+$ that is not a finite ordinal.
But then by definition, $\beta = \gamma^+$ is also a finite ordinal.
Therefore, $\beta$ cannot exist.
In particular, since $\alpha \in \alpha^+$, it follows that:
:$\alpha^+ \subseteq K_I$ implies $\alpha$ is a finite ordinal.
Hence Definition 2 and Definition 3 are equivalent.
Category:Minimally Inductive Set