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It is easy to prove that $\omega_1$ does not have an order-preserving injection into $(\mathbb R, <)$: to each $\alpha \in \omega_1$ we could assign the interval between the image of $\alpha$ and $\alpha + 1$, and this gives uncountably many disjoint open intervals in $\mathbb R$, contradicting its separability.

Intuitively it seems to be that if $\alpha < \omega_1$, there should be an order-preserving injection into $\mathbb R$, but I am having trouble constructing it. If we have $\alpha \to \mathbb R$, we can compose it with a homeomorphism $\mathbb R \to (0, 1)$, and find $\alpha + 1 \to \mathbb R$ by sending $\alpha$ to $(0, 1)$ and $\{\alpha\}$ to 2. However, the limit step is more difficult: if $\lambda < \omega_1$ is a limit, then it is a limit of countably many well-orderings which inject in an order-preserving manner into $\mathbb R$, but you cannot expect these to be "compatible" in a way obvious to me.

Is there a way to make this (or another argument) work? Or are there countable ordinals with no order-preserving injections into $\mathbb R$?

Mees de Vries
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  • You may try this trick for limit ordinals: since for any two ordinals $\alpha, \beta$ one is a subset of another, say, $\alpha < \beta$, you can embed $\alpha$ (an embedding that was already constructed), and then embed $\beta \setminus \alpha$, which is also an ordinal $\leq \beta$. Now, having a sequence $\alpha_1, \alpha_2, \dots$, apply $\mathbb R \rightarrow (0,1)$ to the embedding of $\alpha_1$, then $\mathbb R \rightarrow (1, 2)$ to the embedding of $\alpha_2 \setminus \alpha_1$, $\mathbb R \rightarrow (2, 3)$ to the embedding of $\alpha_3 \setminus \alpha_2$, and so on. – lisyarus Jan 31 '19 at 11:57
  • Do you know the proof of Cantor's theorem about dense countable endless orders ? It's a back-and-forth argument. Well here you can do the same but with just a forth-argument (it works by replacing $\mathbb{R}$ with $\mathbb{Q}$ by the way) – Maxime Ramzi Jan 31 '19 at 12:20
  • I'm sure we've had this before. Every countable total order (not necessarily well ordered) has an order-preserving injection into the rational numbers. – bof Jan 31 '19 at 12:39
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    Countable total orders can be embedded in the rational numbers with a little work, but embedding them in the real numbers is very easy. Let $A$ be a countable totally ordered set. Since it's countable, there is an injection $$g:A\to\left{\frac13,\frac19,\frac1{27},\dots\right}$$. Now the function $$x\to\sum_{y\le x}g(y)$$ is an order-preserving injection from $A$ to $\mathbb R$. – bof Jan 31 '19 at 12:44
  • Thank you for the answers! If any of you want to turn those into an actual answer I'd be happy to accept. Especially the comment that this has nothing to do with the particular order of the countable ordinal is interesting. – Mees de Vries Jan 31 '19 at 13:26

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