Equivalence of Definitions of Minimally Inductive Set

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 infinite successor set.

It is immediate from the definition of a finite ordinal that $\varnothing$ is one.

Also, if $\beta$ is a finite ordinal, so is $\beta^+$.

Therefore:


 * $\omega = \left\{{\alpha: \text{$\alpha$ is a finite ordinal}}\right\}$

is an infinite successor set.

Next, apply Definition 1 to this $\omega$, yielding the set:


 * $\displaystyle \bigcap \left\{{S \subseteq \omega: \text{$S$ is an infinite successor set}}\right\}$

Conversely, suppose that $S \subseteq \omega$ is an infinite successor set.

Suppose that $S \ne \omega$.

Then their difference $\omega \setminus S$ is non-empty.

By Intersection of Ordinals is Smallest:


 * $\alpha := \displaystyle \bigcap \omega \setminus S$

is the smallest element in $\omega \setminus S$.

Then $\alpha \ne \varnothing$, since $\varnothing \in S$ by definition of an infinite successor 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 infinite successor set, this means that:


 * $\beta^+ = \alpha \in S$

which is a contradiction.

It follows that $S = \omega$.

Therefore, we conclude that:


 * $\displaystyle \bigcap \left\{{S \subseteq \omega: \text{$S$ is an infinite successor set}}\right\} = \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 $\alpha = \varnothing$, then $\alpha^+ = \left\{{\varnothing}\right\}$.

Since $\varnothing$ 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 $\alpha \ne \varnothing$, 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 $\beta = \varnothing$ or $\beta = \gamma^+$ for some ordinal $\gamma$.

Since $\varnothing$ 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.