Proof that no uncountable ordinal is a suborder of the reals

Question

I know that every countable ordinal is a suborder of the reals. Is this result sharp? That is, is it the case that no uncountable ordinal is a suborder of the reals?

Answer 1

You're right! Let's show $\omega_1$ doesn't order embed into $\mathbb{R}$. To see why, recall $\mathbb{Q}$ is countable and dense.

Now, if $f : \omega_1 \to \mathbb{R}$ is strictly monotone, we have to have $f(\alpha) < f(\alpha+1)$ for every $\alpha \in \omega_1$. But that means there is a rational between $f(\alpha)$ and $f(\alpha+1)$. Of course, there's only countably many rationals to go around, so we cannot do this.

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I hope this helps ^_^