Apply the same function to both sides of the equation:

\[f_{1+\omega}(n) = f_\omega(n)\]

They're the same. Deal with it.


More specifically, we can look at the definition of the fast-growing hierarchy and the Wainer hierarchy:

  • \(f_0(n) = n + 1\)
  • \(f_{\alpha+1}(n) = f^n_\alpha(n)\), where \(f^n\) denotes function iteration
  • \(f_\alpha(n) = f_{\alpha[n]}(n)\) if and only if \(\alpha\) is a limit ordinal 
  • Where \(\alpha[n]\) is defined as follows:
    • \(\omega[n] = n\)
    • \(\omega^{\alpha + 1}[n] = \omega^\alpha n\)
    • \(\omega^{\alpha}[n] = \omega^{\alpha[n]}\) if and only if \(\alpha\) is a limit ordinal
    • \((\omega^{\alpha_1} + \omega^{\alpha_2} + \cdots + \omega^{\alpha_{k - 1}} + \omega^{\alpha_k})[n] = \omega^{\alpha_1} + \omega^{\alpha_2} + \cdots + \omega^{\alpha_{k - 1}} + \omega^{\alpha_k}[n]\) where \(\alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_{k - 1} \geq \alpha_k\)
    • \(\epsilon_0[0] = 0\) (alternatively \(1\)) and \(\epsilon_0[n + 1] = \omega^{\epsilon_0[n]}\)

We can use this to prove that the supposed inequality discussed on Talk:Ultra-hyper-cosmo-galaxi-cosmo-hyper-ultrol is bull. What's important is the condition that \(\alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_{k - 1} \geq \alpha_k\). We have to express the term in Cantor normal form -- as a finite decreasing series of terms of the form \(\omega^\alpha\) -- before we can apply the \((\omega^{\alpha_1} + \omega^{\alpha_2} + \cdots + \omega^{\alpha_{k - 1}} + \omega^{\alpha_k})[n] = \omega^{\alpha_1} + \omega^{\alpha_2} + \cdots + \omega^{\alpha_{k - 1}} + \omega^{\alpha_k}[n]\) thing.

\[\omega^6+\omega^5+\omega^4+\omega^3+\omega^4+\omega^5+\omega^6 = \omega^6+\omega^6\]

Since \(6 \geq 6\), this is indeed the Cantor normal form of the ordinal, and likewise its fundamental sequence is \(\omega^6+\omega^6[n]\). It would actually be contradictory to the definition of the Wainer hierarchy to say otherwise.

Please, get this through your heads. I'm tired of having to explain this concept this over and over again.

"\(1+\omega\) is not a legal argument"

This is 100% wrong. Let's look at the definition of ordinal addition:

  • \(\alpha + 0 = \alpha\)
  • \(\alpha + S(\beta) = S(\alpha + \beta)\) where \(S\) denotes successor function
  • \(\alpha + \beta = \sup\{\alpha + \gamma | \gamma < \beta\}\) for \(\beta \in \text{Lim}\)

So therefore \(1 + \omega = \sup\{1 + \gamma | \gamma < \omega\} = \sup\{1, 2, 3, 4, \ldots\} = \omega\). \(1 + \omega\) is an ordinal. It's in \(\text{On}\).

In FGH + Wainer hierarchy, \(f_\alpha\) is defined for all \(\alpha \leq \varepsilon_0\). Since \(1 + \omega = \omega\) and \(\omega \leq \varepsilon_0\), \(f_{1+\omega}\) is defined.

The alternative

Okay, so suppose you don't want these equalities. That's fine! But you must define what you're saying and clarify that you're not working with ordinals and FGH, but modified versions. That's good googology.

That's sort of what Sbiis has done, although I object to the fact that he still calls the symbols "ordinals" and he still calls the function "FGH." It's embarrassingly dishonest to introduce something as "FGH" when in fact it is an original variant system, and making little effort to distinguish the two or even introduce the real hierarchy.

In conclusion

As in love as you are with your intuition, it should not override real mathematics. It doesn't matter how you feel FGH should work. It works like this and you're going to have to live with that.

I apologize for the hostile and abrasive tone of this post, but I'm baffled at (and fed up with) the amount of misconceptions around these concepts.

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