User blog comment:TheKing44/Ordinal Definable System of Fundamental Sequences/@comment-35470197-20191201030851/@comment-39605890-20191201175746

> Could you tell me the reason?

Sorry, I did not realize the argument was more involved than I thought.

Okay, let $$g$$ be a system of fundamental sequences. We assume $$\alpha$$ is in $$g$$'s domain, but not odsfs's domain. Moreover, let $$\alpha$$ be the first such ordinal. Also, $$g$$ is defined by the formula $$\phi(g)$$ in the language of first-order set theory.

Now, the formula $$\psi(f,x)=\exists g.(\phi(g) \land f=g(x))$$ defines $$g(\alpha)$$ with $$x$$ assigned to $$\alpha$$. This means that $$g(\alpha)$$ is an ordinal definable fundamental sequence for $$\alpha$$.

Moreover, this implies that $$\alpha$$ has a ordinal definable bijection with the natural numbers. First, we define a surjection from $$\mathbb N \times \mathbb N$$ to $$\alpha$$ as $$f(n_1,n_2)=odsfs(g(\alpha,n_1),n_2)$$. Now this can be used to define a surjection from the naturals to $$\alpha$$, and then a bijection by treating this surjection as a sequence and removing the duplicates. This results in an ordinal definable bijection, contradicting the fact that it is not in odsfs.

A slightly easier way to argue the last paragraph is to say that $$\alpha$$ has cofinality $$\omega$$ according to HOD, and all the ordinals before it are countable according to HOD, and therefore we can argue from within HOD that it is countable since HOD satisfies ZFC.

> Are you stating that the limit \(\alpha\) of a strictly increasing countable sequence \(\alpha_n\) of ordinals admitting ordinal definable fundamental sequences \(\alpha_n[m]\) again admits an ordinal definale fundamental sequences?

That is only the case if that strictly countable sequence is itself ordinal definable.