User blog comment:Plain'N'Simple/A question for proof-theory experts/@comment-35392788-20191029194318/@comment-35392788-20191031133126

I have checked out your blog post, and I have concluded what was so undesirable about it.

The undesirable property of the system you presented is the following : fundamental sequences of some ordinals can contain arbitrarily long subsequences of fundamental sequences of smaller ordinals. More precisely, your system gives \(\omega^{n+1}\) a fundamental sequence that contains all finite ordinals from 0 to n. Since n is obviously unbounded, it's possible to have the fundamental sequence of \(\omega^{n+1}\) contain arbitrarily many elements of the fundamental sequence of \(\omega\).

In formal logic, this property is written as such (greek letters are ordinals and lowercase letters are natural numbers) :

\(\forall n\exists\alpha\exists\beta<\alpha\exists m\geq n\forall k\leq m(\alpha[k]\in\beta)\)

Or, in words : for all n, there exists \(\alpha\) and \(\beta<\alpha\) such that at least n elements of the fundamental sequence of \(\alpha\) belong to \(\beta\).

Its negation, and thus our new condition, is written in formal logic as :

\(\exists n\forall\alpha\forall\beta<\alpha\forall m\geq n\exists k\leq m(\alpha[k]\notin\beta)\)

In words : there exists an n such that no matter the choice of \(\alpha\) and \(\beta<\alpha), at least n elements of the fundamental sequence of \(\alpha\) never all belong to \(\beta\).

This condition thus forbids the system you proposed from being valid. However, this new condition is not a completely new criterion, it's actually a more general case of condition 3. If condition 3 doesn't hold (the fundamental sequence has a smaller ordinal for limit), then trivially arbitrarily many elements of the fundamental sequence belong to a smaller ordinal.