User blog comment:Ikosarakt1/Uncountable function cannot exists/@comment-1605058-20130802172024

About an update - it all depends on choice of \(\alpha_n\). Small lemma - for every n and suffiscently large a we have \(f_a(n)>a\). I will omit a proof, at this is fairly obvious fact. So there is a>f(n) such that \(f_{a}(n)>a>f(n)\). So if we always choose smallest ordinal, it will be finite, so gap also is finite.

Also, as Deedlit pointed out, if we go to large countable ordinals, it all depends on choice of fundamental sequence for every countable ordinal. Any finitely definable ordinal notation names only countably many ordinals (I know there are models of ZFC where everything is definable, but we can use whole Von Neumann universe). Furthermore, extending this argument, there is countable ordinal such that no notation names it. So we may have a problem.