# Epsilon numbers

Let $\alpha$ be an ordinal number and define $f_\alpha$ as:

• $f_\alpha(0) = \alpha + 1$
• $f_\alpha(n+1) = \omega^{f_a(n)}$

Let $S(\alpha) = \sup\{f_a(n)\ |\ n \in \omega\}$

Then $S(\alpha)$ is an epsilon number and is the least epsilon number greater than $\alpha$.

Since none of natural numbers are epsilon number, I think $S(n)=S(m)$ for every natural numbers $n,m$. I know that I'm wrong but I don't know why. Please, help.

And I have problem with showing that $m<n\implies S(m)<S(n)$

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Actually, $S(n)=S(m)$ for all natural numbers $n,m$. Since $\epsilon_0$ is the minimal epsilon number above of them. – azarel Jun 4 '12 at 1:01
Then how come my book says 'for every ordinal number $\alpha & \beta$ , if $\alpha < \beta$ then S($\alpha$) < S($\beta$)? – Katlus Jun 4 '12 at 1:06
Your book probably has lots of typos, white lies, and full blown errors. It is the nature of books. If $\gamma$ is an epsilon number and $\gamma\le\alpha<\beta<S(\gamma)$, then $S(\alpha)=S(\beta)=S(\gamma)$, so you cannot prove the inequality you are asking. – Andrés E. Caicedo Jun 4 '12 at 1:08
Then how can i prove that there exists an isomorphism between 'class of ordinals' and 'class of epsilon numbers'? Or is it false too? – Katlus Jun 4 '12 at 1:12
Oh, that's true. You know that, for any $\alpha$, $S(\alpha)>\alpha$ and $S(\alpha)$ is epsilon. So, you can enumerate the epsilons in increasing order: $\epsilon_0=S(0)$. Given $\epsilon_\alpha$, let $\epsilon_{\alpha+1}=S(\epsilon_\alpha)$. And for limit $\lambda$, let $\epsilon_\lambda=\sup\{\epsilon_\alpha\mid\alpha<\lambda\}$. You need to check: 1. Every $\epsilon_\alpha$ is an epsilon. This is clear except if $\alpha$ is limit. 2. Every epsilon is an $\epsilon_\alpha$. This can be proved by considering a least putative counterexample. – Andrés E. Caicedo Jun 4 '12 at 1:18

Actually, $S(n)=S(m)$ for all natural numbers $n,m$. Since $\epsilon_0$ is the minimal epsilon number above of them.

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