Equivalence of Definitions of Minimal Infinite Successor Set

Theorem

The following definitions of the concept of Minimal Infinite Successor Set are equivalent:

Definition 1

Let $S$ be an infinite successor set.

The minimal infinite successor set $\omega$ is the infinite successor set given by:

$\ds \omega := \bigcap \set {S' \subseteq S: S' \text{ is an infinite successor set} }$

that is, $\omega$ is the intersection of every infinite successor set which is a subset of $S$.

Definition 2

The minimal infinite successor set $\omega$ is defined as the set of all finite ordinals:

$\omega := \set {\alpha: \alpha \text{ is a finite ordinal} }$

Definition 3

The minimal infinite successor set $\omega$ is defined as:

$\omega := \set {x \in \On: \paren {x \cup \set x} \subseteq K_I}$

where:

$K_I$ is the class of all non-limit ordinals
$\On$ is the class of all ordinals.

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 infinite successor 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 infinite successor set.

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

$\ds \bigcap \set {S \subseteq \omega: S \text{ is an infinite successor set} }$

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.

$\ds \alpha := \bigcap \omega \setminus S$

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

Then $\alpha \ne \O$, since $\O \in S$ by definition of an 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:

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

It follows that $S = \omega$.

Therefore, we conclude that:

$\ds \bigcap \set {S \subseteq \omega: S \text{ is an infinite successor set} } = \omega$

establishing the equivalence of the definitions.

$\Box$

Definition 2 equals Definition 3

From Definition 2 and Definition 3, we see that we are to prove:

$\alpha$ is a finite ordinal if and only if $\alpha^+ \subseteq K_I$

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.

$\blacksquare$