# Second element of fundamental sequence of Zeta-nought

I'm attempting to evaluate the second element of the fundamental sequence for the infinite ordinal known as zeta_0

zera_0, is defined as the supremum of: {epsilon_0, epsilon_(epsilon_0), epsilon_(epsilon_(epsilon_0)), ...}

where epsilon_(n+1) is defined as the supremum of: {epsilon_n, epsilon_n^epsilon_n, epsilon_n^epsilon_n^epsilon_n, ...}

epsilon_0, in turn is defined as the supremum of: {omega, omega^omega, omega^omega^omega, ...}

and omega is the infinite ordinal which is the supremum of the natural numbers: {1, 2, 3, ...}

I've given the evaluation of the second element of this ordinal's fundamental sequence (for the purposes of a theoretical evaluation of the fast growing diagonalization hierarchy) an attempt and here is what I have so far.

zeta_nought = sup{epsilon_0, epsilon_(epsilon_0), ...} epsilon_(epsilon_0) = sup{epsilon_omega, epsilon_(omega^omega), ...} skipping a few easy ones, we get to: epsilon_(omega+2) = sup{epsilon_(omega+1), epsilon_(omega+1)^epsilon_(omega+1), ...} epsilon_(omega+1)^epsilon_(omega+1) = sup{epsilon_(omega+1)^epsilon_omega, epsilon_(omega+1)^epsilon_omega^epsilon_omega, ...} epsilon_(omega+1)^epsilon_omega^epsilon_omega = sup{epsilon_(omega+1)^epsilon_omega^epsilon_0, epsilon_(omega+1)^epsilon_omega^epsilon_1, ...} epsilon_(omega+1)^epsilon_omega^epsilon_1 which gives epsilon_(omega+1)^epsilon_omega^epsilon_0^epsilon_0 which gives epsilon_(omega+1)^epsilon_omega^epsilon_0^omega^omega which gives epsilon_(omega+1)^epsilon_omega^epsilon_0^(omega+2)

This is where I'm stuck.

I have made several attempts to move on from this point. One of which lead to this mess:

epsilon_(omega+1)^epsilon_omega^(omega^4 + 8*omega^3 + 24*omega^2 + 32*omega + 16)

But I feel like I made several bad assumptions and missteps to arrive at that. At one point along the way, I took epsilon_0^(omega+2) and turned it into epsilon_0^omega * epsilon_0^2, then evaluated the first factor to epsilon_0^2, and turned the whole thing into epsilon_0^4.

This felt wrong and I'm pretty sure it is.

Please let me know if there's anything I can do here to make this question more readable. I'm still not familiar with the markup syntax here.

• Wouldn't it be simpler to define $\zeta_0$ to be the least fixed point in the $\varepsilon$ enumeration, i.e. the least $\alpha$ such that $\alpha=\varepsilon_\alpha$, where $\varepsilon_\alpha$ are the usual $\varepsilon$ numbers? – Asaf Karagila Aug 19 '13 at 21:47
• Also, this is really impossible to understand properly without using $\LaTeX$. – Asaf Karagila Aug 19 '13 at 21:48
• I don't think your formulation will aid in the evaluation of the second element of the fundamental sequence though. I did want to try to make this prettier, but I wasn't sure how to use subscripts/superscripts, and greek letters. – David Bandel Aug 19 '13 at 22:00
• The phrase "the fundamental sequence" in the question suggests that you're working in a context where someone has chosen a particular fundamental sequence for each ordinal (or at least for ordinals of the size you're interested in). I'm not aware that any particular choice is universally accepted. If it were up to me, I'd choose the fundamental sequence for $\zeta_0$ to be the $\varepsilon$-iterations as in the question. (There would still be a question about the meaning of "second term", i.e., whether you start with a $0$-th term or not.) – Andreas Blass Aug 19 '13 at 22:12
• meta.math.stackexchange.com/questions/5020/… might help you. – Asaf Karagila Aug 19 '13 at 22:14

A fundamental sequence for a limit ordinal is simply an increasing sequence of ordinals whose supremum is the aforementioned ordinal. Unfortunately, there is not a single definition of fundamental sequence for all ordinals, but for $\zeta_0$ it would be natural to set

$\zeta_0 [0] = 0$

$\zeta_0 [n+1] = \varepsilon_{\zeta_0 [n]}$

So $\zeta_0 [2]$ would be $\varepsilon_{\varepsilon_0}$. However, it seems like what you want is to keep taking the second element of the fundamental sequence until you get to a successor ordinal. This can be done, but you need to make sure that you have predefined your fundamental sequences in a consistent way. For each ordinal less than or equal to $\zeta_0$, you want a single fundamental sequence, so you need to be sure that two different expressions for the same ordinal don't generate two different fundamental sequences. Towards that end, it is helpful to have a canonical form for every ordinal up to $\zeta_0$.

Here is one way to define a canonical form: For each ordinal $\alpha$, write it as a sum of additively principal ordinals (that is, powers of $\omega$): $\alpha_1 + \alpha_2 + \ldots + \alpha_n$. If $\alpha$ is additively principal but not an $\varepsilon$-number, write it as $\omega^\beta$. If $\alpha$ is an $\varepsilon$-number but not a $\zeta$-number, write it as $\varepsilon_\beta$. If $\alpha$ is a $\zeta$-number, write it as $\zeta_\beta$. That will give you a single expression for each ordinal, so you can define fundamental sequences consistently.

Your fundamental sequence values seem a little inconsistent; It seems that you are using the rule $\varepsilon_\alpha [n] = \varepsilon_{\alpha[n]}$, which is a good rule, but then you have $\omega[2] = 2$ but $\varepsilon_\omega [2] = \varepsilon_1$, which seems inconsistent. I would set $\varepsilon_\omega [2] = \varepsilon_2$.

Other than that, your evaluations seem fine, but you were right that you made a mistake but setting $\varepsilon_0^{\omega + 2} [n] = \varepsilon_0^{n+2}$. The problem is that $\varepsilon_0^{n+2}$ has limit $\varepsilon_0^{\omega}$, not $\varepsilon_0^{\omega + 2}$. In general, the function $f(\alpha) = \alpha \beta$ is not continuous, so setting $(\alpha \beta) [n] = (\alpha [n]) \beta$ does not work in general. The function $f(\beta) = \alpha \beta$ is continuous, so setting $(\alpha \beta) [n] = \alpha (\beta[n])$ will work. However, remember to express each ordinal in a canonical form and define a specific set of rules for your fundamental sequences.