Can we get uncountable ordinal numbers through constructive method?

Question

As we know, $2^{\aleph_0}$ is a limit ordinal number, however, it is greater than $\omega$, $\omega+\omega$, $\omega \cdot \omega$, $\omega^\omega$, $\omega\uparrow\uparrow\omega$, and even $\omega \uparrow^{\omega} \omega$.

My question is can we get uncountable ordinal numbers with only natural number, $\omega$ and ordinal hyperoperation through constructive method?

Answer 1

There is a trivial way, namely $\omega$ and $1$ and the set theoretical operation "the next initial ordinal" which gives us $\omega_1$.

For other operations, we can do the following claim:

Suppose $\ast$ is an ordinal operation such that for $\alpha,\beta$ countable we have $\alpha\ast\beta$ is countable, then every countable ordinal $\gamma$, the $\gamma$-th iteration, $\alpha\ast^\gamma\beta$ is countable.

We need to define exactly what does iterations mean:

• $\alpha\ast^{\gamma+1}\beta=(\alpha\ast^\gamma\beta)\ast\beta$;
• If $\delta$ is a limit ordinal then $\alpha\ast^\delta\beta=\sup\{\alpha\ast^\gamma\beta\mid\gamma\lt\delta\}$.

The proof is by induction over the iterations:

1. We already assume that if $\alpha$ and $\beta$ are countable then $\alpha\ast\beta$ is countable.

2. Suppose that for iterations of length $\gamma$ we know that $\alpha\ast^\gamma\beta$ is countable, then $\alpha\ast^{\gamma+1}\beta$ is countable since we apply $\ast$ to two countable ordinals.

3. If $\delta$ is a limit ordinal, and $\alpha\ast^\gamma\beta$ is countable for all $\gamma<\delta$ then we have that $\alpha\ast^\delta\beta$ is a countable limit of countable ordinals and therefore countable.