User blog comment:Eners49/A whole new superclass of infinities?/@comment-35470197-20180722021627/@comment-35470197-20180723235116

@Syst3ms

Uh. Ok. I am very sorry for that. I really thought that the OP was using set theory (with several misunderstandings on infinity), because the OP was mentioning ordinals such as \(\varphi(4,0)\), which are much more difficult than elementary ordinals easily described in arithmetic such as \(\omega\) and \(\omega^{\omega}\).

Thank you for pointing out that very few googologists are familiar with set theory. I did not know the fact at all, because many googologists use the PTO of KP set theory and also I was suggested to use NBG set theory instead of ZFC set theory in my blog post. Since even mathematicians do not necessarily know KP set theory or NBG set theory unless they are logicians or categorists, I thought that many googologists here are unbelieavably familiar with set theories.

@PsiCubed2

I see. Thank you for telling me the method to explain in a better manner. I will try explanations in the natural language in such replies as far as possible. Your explanation is outstandingly sophisticated (pretty easy to read and understand!), and I hope that I will be able to explain things as you do.

> You cannot have a "largest ordinal" for the exact same reason that you cannot have a "largest (finite) number": You can always add 1.

Yeah. You are completely right. I would like to add why I wrote this. I though that the OP was considering the property like \(\infty + 1 = \infty\) as in the real analysis (without contradiction). Then I guessed that the "greatest ordinal" does not actually mean an ordinal, but means an object "beyond" all ordinals in this context.

So what I explained is Burali-Forti theorem, which states that classes containing all ordinals are never sets.

@user

I guess that you are referring only to recursive ordinals provable under ZFC.