User blog:Ikosarakt1/Uncountable function cannot exists

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 largest n such that \(f_\alpha(n) > f(n)\). 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.