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I've been learning about and playing around with ordinals, and I've come across the following question:

Suppose we take a subset of the real numbers $S\subseteq\mathbb{R},$ under the ordering assigned to $(\mathbb{R},\le),$ such that this subset is well-ordered. Each well-ordered set can be assigned an ordinal; what's the smallest ordinal that can't be assigned to a well-ordered set?

Things I've found so far (I'll be calling an ordinal "real-orderable" if it can be assigned to a well-ordered set):

  • If $\alpha$ is real-orderable, $\alpha+1$ is as well, since we can compress the set corresponding to $\alpha$ into $(0,1)$ and tack another element onto the end. (This is why I asked for the smallest impossible rather than the largest possible.)
  • If $\alpha$ is real-orderable, $\omega\alpha$ is as well, since we can take (countably) infinite copies of the compressed set and put them on all $(n,n+1),$ $n\in\mathbb{N}.$
  • The first uncountable ordinal $\omega_1$ is not real-orderable, since well-ordered sets can assign a rational number within the following interval to every element.
  • I suspect that if $\alpha$ is orderable, $\omega^\alpha$ is as well, but I can't find a nice way to prove it.

My best guesses are (in order of likelihood) $\varepsilon_0,$ $\omega_1,$ $\omega_1^{\text{CK}},$ $\omega^\omega.$ Is there a good way to approach this problem?

Bob Translator
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    The ordinals embeddable into $\mathbb{R}$ are exactly the countable ordinals, so the answer is $\omega_1$. – Qiaochu Yuan Nov 02 '22 at 04:52
  • For previous math.SE discussions of this see: https://math.stackexchange.com/questions/408300/countable-ordinals-are-embeddable-in-the-rationals-bbb-q-proofs-and-their, https://math.stackexchange.com/questions/44559/formal-proof-for-a-subset-of-the-real-numbers-well-ordered-with-the-normal-orde, https://math.stackexchange.com/questions/123969/embedding-ordinals-in-mathbbq – Qiaochu Yuan Nov 02 '22 at 05:56

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