=== 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. {{qed|lemma}} === Definition 2 equals Definition 3 === From Definition 2 and Definition 3, we see that we are to prove: :$\alpha$ is a finite ordinal {{iff}} $\alpha^+ \subseteq K_I$ 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. {{qed}} Category:Minimal Infinite Successor Set pv6ifpxhjnuo9e8gi2zv6k5xkbzub3c