User blog comment:Vel!/Music/@comment-5150073-20140503054608/@comment-1605058-20140505132646

@FBZ CH isn't necessary, but axiom of choice is required (not fully, but dependent choice isn't enough).

@Iko I have no idea how injection from $$\omega_1$$ to $$\mathbb{R}$$, even order-preserving one, could give us $$f_{\omega_1}$$.

Also, your argument using CH is wrong, as even if there is as many functions $$\mathbb{N}\rightarrow\mathbb{N}$$ as countable ordinals, it doesn't tell us that ALL such functions would be used up by FGH. It's similar to saying that, since there is the same number of even and natural numbers, there can be no odd number.

Argument using $$2^{\aleph_1}$$ also fails, but for a different reason - just note that there is exactly as many functions which take values in interval $$(0,1)$$, and none of these is reasonable definition for $$f_{\omega_1}$$.

@FBZ2 Even stronger statement can be made: There is no way (nice or not) to define FGH all the way below $$\omega_1$$. The reason for this is following: every ordinal below $$\omega_1$$ has a fundamental sequence, but if we wanted to define it for every countable ordinal, we would require an obvious application of axiom of choice. Non-obvious part is that it cannot be done without choice. It's a fact that if proof of existence of an object can be found in ZFC but not ZF, then there is no constructible (in natural meaning of this word) example of it. In particular, result in bold follows.

As this comment is already long, I'll point out one things: it's possible for $$\omega_1$$ to have fundamental sequence. The point is that "limit of countably many countable ordinals is countable" is not provable in ZF alone. We need a tad bit of choice to achieve it (it follows from countable choice).