Are there ordinals beyond all the $\omega$'s?

Question

Are there ordinals that are somehow "beyond" all the $\omega$'s?

Answer 1

Well. Just off the bat, $\omega_1$ is an ordinal, and it cannot be expressed as a "polynomial" in $\omega$ in any nontrivial way.

Moreover, since for infinite ordinals, $\alpha,\beta$ we have that $|\alpha+\beta|=|\alpha\cdot\beta|=|\alpha^\beta|=\max\{|\alpha|,|\beta|\}$ (where the arithmetic is ordinal arithmetic, of course), it follows that any finite expression involving countable ordinals is countable. So there is some countable ordinal which cannot be expressed in such nontrivial way.

As luck would have it, we also know that this ordinal is $\varepsilon_0$, and can be defined as the least ordinal $\alpha$ such that $\omega^\alpha=\alpha$, or as $\sup\{\omega,\omega^\omega,\omega^{\omega^\omega},\ldots\}$.

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Of course, if initial ordinals $\omega_\alpha$ are considered in your question as "$\omega$'s" then the answer is negative. Hartogs theorem ensures that if $X$ is a set (e.g. an ordinal), then there is an infinite ordinal which does not inject into $X$, and the least such ordinal is some $\omega_\alpha$ for some $\alpha$.