User blog comment:Simplicityaboveall/Extremely Large Numbers 4/@comment-5529393-20160801195700/@comment-5529393-20160803061403

Simplicity,

You can look up the history of the SGH page, it looks like Ikosarakt1 did much of the comparisons, and Googleaarex did some as well. One must keep in mind, though, that the SGH is very sensitive to changes in the fundamental sequences used to define it. The comparisons given on the page may well be valid for a particular set of fundamental sequences; however I don't think they have been carefully vetted or anything like that, so reader beware.

The problem with the "decomposition function" is that, much like the other terms that we have been talking about, it is not well defined. You just say that "every ordinal can be decomposed into omega", and that's it, no description of how such a thing is accomplished. And that statement is not true; every ordinal below omega_1 cannot be described with distinct notations, due to the uncountability of the countable ordinals; similarly, every ordinal below omega_1^CK cannot be described with a computable notation.

It seems like you have faith in the notion that nothing needs to be done to create this "decomposition function", that every ordinal has a natural, unique expression in terms of omega. This is simply not true.

I can try to help you with the R function. What we need is an algorithm that takes any natural number, and uniquely creates an expression for it in terms of 10. That expression must have meaning when we replace 10 with omega, and hopefully the expressions with omega go up to a fairly large ordinal.

Here's one idea: the down-arrow notation.

$$ a \downarrow c = a^c $$

$$ a \downarrow^b 1 = a $$

$$ a \downarrow^{b+1} (c+1) = (a \downarrow^{b+1} c) \downarrow^b a $$

This notation works differently from uparrows:

$$ a \downarrow^2 c = a^{a^{c-1}} $$

$$ a \uparrow^2 c < a \downarrow^3 c < a \downarrow^4 c < a \uparrow^3 c $$

$$ a \uparrow^3 c < a \downarrow^5 c < a \downarrow^6 c < a \uparrow^4 c $$

so it's a little unusual, but the advantage is that it works much more nicely with transfinite ordinals than up-arrows do.

Here is the definition for general ordinals:

$$ \alpha \downarrow \gamma = \alpha^\gamma$$

$$ \alpha \downarrow^\beta 1 = \alpha $$

$$ \alpha \downarrow^{\beta+1} (\gamma+1) = (\alpha \downarrow^{\beta+1} \gamma) \downarrow^{\beta} \alpha $$

If $$\beta$$ is a limit ordinal, $$ \alpha \downarrow^\beta (\gamma+1) = \sup_{\delta < \beta} \lbrace (\alpha \downarrow^\beta \gamma) \downarrow^\delta \alpha \rbrace $$

If $$\gamma$$ is a limit ordinal, $$ \alpha \downarrow^\beta \gamma = \sup_{\delta < \gamma} \lbrace \alpha \downarrow^\beta \delta \rbrace $$

Okay, so now we have a notation that gives natural numbers when we plug in natural numbers, and gives transfinite ordinals when we plug in transfinite ordinals. Next, we need to uniquely define a notation given a number m and a base n.

For are first step, choose a_1 to be maximal such that $$ 10 \downarrow^{a_1} 2 \le m$$. Then choose b_1 to be maximal such that $$ 10 \downarrow^{a_1} b_1 \le m$$ Next, subtract $$10 \downarrow^{a_1}b_1$$ from m, and then choose a_2 and b_2 to be maximal such that $$ 10 \downarrow^{a_2} b_2 \le m - 10 \downarrow^{a_1} b_1$$. Continue on in this fashion, until we reduce the remainder to a number less than 100. Express the remainder as a sum of 10's plus a number less than 10. So

$$m = 10 \downarrow^{a_1} b_1 + 10 \downarrow^{a_2}b_2 + \ldots + 10\downarrow^{a_n}b_n + 10m + r$$

Then apply the same procedure to all the a_i and b_i, until everything is written in terms of 10 and numbers less than 10. Now we can replace 10 with omega, and we have our R function.

One more thing we have left to do, which I forgot to mention that you needed: we need to define fundamental sequences for our FGH! In your articles you define fundamental sequences for successor epsilon numbers, but we need to be more general. So here goes:

If our expression for $$\alpha$$ ends in $$\omega \cdot m$$, then $$\alpha[k] = \omega\cdot (m-1) + k$$

If our expression for $$\alpha$$ ends in $$\omega \downarrow^{\alpha_n} \beta_n$$ where $$\beta_n$$ is a limit ordinal, then

$$ \alpha[k] = \omega\downarrow^{\alpha_1}\beta_1 + \ldots + \omega \downarrow^{\alpha_n} (\beta_n[k])$$

If our expression for $$\alpha$$ ends in $$\omega \downarrow^{\alpha_n}\beta_n$$ where $$\alpha_n$$ is a limit ordinal and $$\beta_n = \gamma + 1$$, then

$$ \alpha[k] = \omega\downarrow^{\alpha_1}\beta_1 + \ldots + (\omega \downarrow^{\alpha_n} \gamma) \downarrow^{\alpha_n[k]} \omega $$

If our expression for $$\alpha$$ ends in $$\omega \downarrow^{\alpha_n}\beta_n$$ where $$\alpha_n = \gamma+1$$ and $$\beta_n = \delta + 1$$, then

$$ \alpha[k] = \omega\downarrow^{\alpha_1}\beta_1 + \ldots + (\omega \downarrow^{\alpha_n} \delta) \downarrow^{\gamma} k $$

and that's it.

So, this gives you a well-defined notation system. However, it only goes to $$\Gamma_0$$. You can go through the same process for linear arrays, and that will get you up to the Small Veblen Ordinal. A stronger notation is to use {a1 @ b1, a2 @ b2, ..., an @ bn}, where ai @ bi means that the ordinal ai is in the bi'th ordinal place. This will get you up the Large Veblen Ordinal. If you want stronger ordinals - well, it will be tricky, since you need a notation that not only generates large ordinals in terms of omega, but gets you finite ordinals when you replace omega by 10.

Welp, good luck!