Why aren't there uncountably many disjoint open intervals of $\mathbb{R}$?

Question

I know that this can't be true given that $\mathbb{R}$ is separable, but I'm having a hard time coming to grips with why this is, exactly. In particular, can't I just take some uncountable strictly increasing sequence of elements of $\mathbb{R}$, $\langle x_\alpha\mid \alpha<\omega_1\rangle$, and get uncountably many disjoint open intervals $(x_\alpha, x_{\alpha+1})$? I clearly have some kind of misunderstanding, but I'm having a hard time isolating it.

Answer 1

You can’t do what you suggest, because there is no such sequence in $\Bbb R$. If there were, the intervals $(x_\alpha,x_{\alpha+1})$ would indeed be pairwise disjoint and non-empty, so each would contain a rational number; say $q_\alpha\in(x_\alpha,x_{\alpha+1})\cap\Bbb Q$. We now have a function $\varphi:\omega_1\to\Bbb Q:\alpha\mapsto q_\alpha$. Since $\Bbb Q$ is countable and $\omega_1$ is not, this map cannot be injective, and there must therefore be $\alpha<\beta<\omega_1$ such that $q_\alpha=q_\beta$. But then $q_\alpha\in(x_\alpha,x_{\alpha+1})\cap(x_\beta,x_{\beta+1})=\varnothing$, which is absurd.

In other words, the argument that the separability of $\Bbb R$ implies the non-existence of uncountably many pairwise disjoint, non-empty open intervals is actually a proof that no such strictly increasing $\omega_1$-sequence of reals exists. Equivalently, it shows that if $\langle x_\alpha:\alpha<\omega_1\rangle$ is a weakly increasing $\omega_1$-sequence of reals, then there must be an $\eta<\omega_1$ such that $x_\alpha=x_\eta$ for all $\alpha$ such that $\eta\le\alpha<\omega_1$.

Answer 2

Well, as you seem to understand, the proof that any collection of disjoint open intervals is countable is very simple. Also it's correct.

On the other hand, if you start with a strictly increasing "sequence" $\langle x_\alpha\mid \alpha<\omega_1\rangle$ then yes, it follows that $(x_\alpha,x_{\alpha+1})$ is an uncountable family of disjoint open intervals. So why can't you do that? Evidently because there is no such sequence $\langle x_\alpha\mid \alpha<\omega_1\rangle$. We've proved that.