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Prove that if an ordinal $\alpha$ can be embedded (in an order preserving manner) into $\Bbb{R}$, then $\alpha$ is countable.

I managed to prove this using $\mathsf{AC}$ (countable choice is sufficient), in two ways:

Proof 1. Assume $f:\alpha\to\Bbb{R}$ is order-preseving. Let $S$ be the set of all half-open intervals of form $\left[f(\beta),f(\beta+1)\right)$ for $\beta<\alpha$. Obviously, every such interval corresponds to a unique ordinal $\beta < \alpha$ by taking its minimum, therefore $|\alpha| = |S|$. As each interval is non-empty, it contains at least one rational number; by $\mathsf{AC}$ there's a choice function $g:S\to \Bbb{Q}$, and it has to be injective. That proves $S$ is (at most) countable, and so is $\alpha$.

Proof 2. Again, let$f:\alpha\to\Bbb{R}$ be order-preseving. Using any strictly monotone bijection $\Bbb{R}\to (0,1)$ we can assume $f[\alpha]\subseteq (0,1)$. We can obviously embed $\alpha +1$ in $(0,1]$ by sending $\alpha$ to $1$, so assume w.l.o.g. that $\alpha = \beta^+$ is a succesor ordinal, this is to use the fact that $\sup(\alpha) = \beta$. We'll now show that $A=f[\alpha]$ is sequentially closed. Let $(a_n)_{n<\omega}$ be a convergent series of elements of $A$. $(a_n)_{n<\omega}^\infty$ has a monotone subseries, $(b_n)_{n<\omega}$. If $(b_n)_{n<\omega}$ is non-increasing, since it is bounded in $(0,1)$, it converges to its infimum; but $(A,<)$ is a well-order, so the limit is $\min\{b_n\mid n<\omega\}\in A$. Otherwise, if $(b_n)_{n<\omega}$ is non-decreasing, it converges to its supremum; but $\alpha$ is closed under supremum, and therefore $A$ must be too. So, $A$ is sequentially closed, and thus, using $\mathsf{AC}$, closed. It is therefore a countable union of closed intervals. If at least one interval is non-degenarate, it contains an open sub-interval, which does not admit a minimum, contrary to the assumption that $(A,<)$ is a well-order. Therefore, $A$ is at most countable.

My questions are:

  1. Are these proofs valid (especially the second one)?
  2. Is it possible to prove it without $\mathsf{AC}$?
  3. This theorem seems closely related to Cantor-Bendixson analysis and perfect sets. Is there any proof that involves such tools?
  4. Is there some order-embedding version of Hartogs' number of a total order $(X,<)$? I.e., minimal ordinal $\alpha$ s.t. $\alpha$ cannot be embedded into $X$. Can one define it to be, analogously to Hartogs' number, the set of all ordinals that can be embedded into $X$?

Related but different questions include this question and this question.

Roy Sht
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    I don't get the second proof. The image $f[\alpha]$ is not necessarily closed, because $f$ is not assumed to be continuous. Every countable ordinal is order-isomorphic to a discrete subset of $\mathbb R$. – bof Jul 09 '22 at 23:17
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    You don't need AC to choose rational numbers; just fix an enumeration of rational numbers, and whenever you need to choose a rational number take the one with lowest index. – bof Jul 09 '22 at 23:19
  • @bof Thank you, I changed the explanation. I think $A=f[\alpha]$ must still be closed, even if $f$ is not continuous. If not, where is my mistake? – Roy Sht Jul 10 '22 at 00:22
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    @RoySht Consider $$f:\omega+1\rightarrow\mathbb{R}: x\mapsto\begin{cases} 1-{1\over n+1} & \mbox{ if } n<\omega\ 17 & \mbox{ if }n=\omega.\end{cases}$$ We have $1\in cl(f[\omega+1])\setminus f[\omega+1]$. – Noah Schweber Jul 10 '22 at 00:40
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    If $\alpha=\omega+1$ and $f(n)=\frac n{2n+1}$ for $n\lt\omega$ and $f(\omega)=1$, then $f$ is order-preserving but $f[\alpha]$ is not closed. I think your mistake is where you write "and therefore $A$ must be too." – bof Jul 10 '22 at 01:06

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