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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?

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What do you mean constructive? Also,… – Asaf Karagila Jun 7 '12 at 11:16
Cantor called $\omega_1$ the "second number class" since there is a new method of construction required. Up to that point, the two operations "add 1" and "sup of a (countable) sequence)" suffice to generate all ordinals. But they do not suffice for $\omega_1$. – GEdgar Jun 7 '12 at 12:11
@GEdgar: If you do "add $1$" and "$\sup$" finitely many times you don't even get to $\epsilon_0$... – Asaf Karagila Jun 7 '12 at 12:23
@Asaf: that is why I did not say "finitely many times". – GEdgar Jun 7 '12 at 12:27
up vote 2 down vote accepted

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.

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In the third step, do thou need to assume $\delta$ is also countable? It seems if $\delta$ is uncountable, then $\{\alpha *^{\gamma} \beta|\gamma<\delta\}$ would contain uncountable many elements. By the way, if $\delta$ is countable, then $\alpha*^{\delta}\beta$ indeed countable. – Popopo Jun 7 '12 at 12:20
@Popopo: Of course that $\delta$ is countable. The claim only says that for a countable ordinal the result is countable. – Asaf Karagila Jun 7 '12 at 12:21

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