Ordinals closed under functions

Question

Let $ \{ f_n : n \in \mathbb N \} $ be a set of functions $f : (\omega_1)^k\to \omega_1 $ where the $k$ is different between functions. Prove that the set of ordinals $\alpha < \omega _1 $ that are closed with respect to all the functions $f_n$, is unbounded and closed.

Answer 1

Show that for a given ordinal $\alpha$, there is only a countable set of possible values all the $f_n$'s can take on $\alpha^n$, take the supremum of these, and reiterate $\omega$ times. Show that the supremum of these $\omega$ ordinals is closed under all the functions.

Use the above technique to show that the set of ordinals which is closed under the functions is unbounded. The closure is simpler, if $\alpha$ is the limit of $\alpha_n$ where each $\alpha_n$ is closed under the functions, then whenever $\vec\beta$ is a $k$-tuple of ordinals below $\alpha$, there is some $\alpha_n$ which bounds them and therefore $f_i(\vec\beta)<\alpha_n<\alpha$.

Answer 2

Proving that the set of ordinals at which all of the functions are closed is itself closed is very straightforward; the harder part is showing that it’s unbounded. You can do this all at once. Let $\alpha_0<\omega_1$ be arbitrary. Let $\{\langle m_k,n_k\rangle:k\in\omega\}$ be an enumeration of $\omega\times\omega$. Let

$$\alpha_{k+1}=\alpha_k\cup\sup\left\{f_{m_k}(\overline\beta):\overline\beta\in\alpha_k^{r_k}\right\}\;,$$

where $r_k$ is the arity of $f_{m_k}$. Show that $\sup_{k\in\omega}\alpha_k$ is closed under all of the functions.