Equivalence of Definitions of Minimally Inductive Set

Theorem
The following definitions of the minimal infinite successor set are equivalent:

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 $\omega$.

It follows that the set specified by Definition 1 is a subset of that of Definition 2.

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

Definition 1 equals Definition 3
We will prove that both definitions of $\omega$ specify the same set.

Firstly, let us show that $\omega$ as given in Definition 3 is an infinite successor set.

By Empty Set is Ordinal, $\varnothing$ is an ordinal.

By Successor Set of Ordinal is Ordinal, it follows that $\varnothing \in K_I$.

Subsequently, we have that $\varnothing \cup \left\{{\varnothing}\right\} \subseteq K_I$, so that $\varnothing \in \omega$.

Now suppose that $x \in \omega$.

Again by Successor Set of Ordinal is Ordinal, we have that $x^+$ is an ordinal.

The assumption that $x \in \omega$ means that $x^+ \subseteq K_I$.

It only remains to be shown that $x^+ \in K_I$.

However, $x^+$, being a successor ordinal, is by definition a non-limit ordinal.

Thus $x^+ \in K_I$, and we have that:


 * $x^+ \cup \left\{{x^+}\right\} \subseteq K_I$

That is, $x^+ \in \omega$.

Hence, $\omega$ is an infinite successor set.

Next, apply Definition 1 to this $\omega$.

It follows that the set specified by Definition 1 is a subset of that of Definition 3.

Conversely, if we can prove that the $\omega$ from Definition 3 is a subset of any infinite successor set, then in particular:


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

so that the set specified by Definition 3 is a subset of that of Definition 1.

Suppose there were some $x \in \omega$ such that $x \notin \displaystyle \bigcap \left\{{S' \subseteq S: \text{$S'$ is an infinite successor set}}\right\}$.

By definition of $\omega$, it follows that $x$ is a non-limit ordinal.

By Successor Set of Ordinal is Ordinal, $x^+$ is also an ordinal.

Moreover:


 * $\left\{{y \in x^+: y \notin \displaystyle \bigcap \left\{{S' \subseteq S: \text{$S'$ is an infinite successor set}}\right\}}\right\}$

is non-empty, and hence has a smallest element $y$.

Since $x^+ \subseteq K_I$, it follows that $y$ is a non-limit ordinal as well.

Hence $y = \varnothing$ or $y = z^+$ for some ordinal $z$.

The former case is impossible since $\varnothing$ is part of every infinite successor set.

In the latter case, the definition of $y$ as the smallest element of:


 * $\left\{{y \in x^+: y \notin \displaystyle \bigcap \left\{{S' \subseteq S: \text{$S'$ is an infinite successor set}}\right\}}\right\}$

means that:


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

But by Intersection of Infinite Successor Sets, $\displaystyle \bigcap \left\{{S' \subseteq S: \text{$S'$ is an infinite successor set}}\right\}$ is an infinite successor set.

Therefore, we have that:


 * $y = z^+ \in \displaystyle \bigcap \left\{{S' \subseteq S: \text{$S'$ is an infinite successor set}}\right\}$

contradicting the existence of $y$.

It follows that for all $x \in \omega$:


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

that is, the set specified by Definition 1 is a subset of that of Definition 3.

Hence by definition of set equality, the two are equal, and the definitions equivalent.