## FANDOM

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Last days, I bothered to find some function f(n) that will be on par with hypothetically definable $$f_{\omega_1}(n)$$. By the definition $$\omega_1$$ can't has any fundamental sequence. I propose the following way to determine an order type (called: $$\beta$$) for the function f(n):

Define f-typed complexity of ordinal $$\alpha$$ as the smallest natural n such that $$f_\alpha(n) < f(n)$$. If it doesn't exists, then the complexity is 0. Then the n-th term of fundamental sequence for the ordinal $$\beta$$ will be as follows:

If any ordinals with complexity n exist, then the term is the largest of them.

If none of them exist, then the term is the previous term+1.

So, for any f(n) we can construct the countable ordinal (as it has a fundamental sequence) measuring its order type. It can lead to somewhat strange result: I(n) can't diagonalize over all countable ordinals. Even if we define E(n) to be the largest number expressable with n English words (and make it well-defined eliminating inconsistent definitions), still: the ordinals will be countable.

### Update 02.08.2013

Suppose there is a function f(n) which outgrows each countable ordinal in FGH. For each n there will be an ordinal $$\alpha_n$$ so that $$f_{\alpha_n}(n)>f(n)$$. So, there is a question: how big is a gap between two consecutive ordinals: $$\alpha_m$$ and $$\alpha_{m+1}$$. Since they are all countable, then this gap will be countable too, and we can have only countable number of this ordinals (actually: n of them).