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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.

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  • $\begingroup$ Any open interval in $\mathbb{R}$ contains a rational number (I mean in the usual Euclidean topology). Take that rational number and label that open interval with it. Doing the same for any collection of disjoint open intervals you get a subset of $\mathbb{Q}$ which is countable. $\endgroup$ – Hamed May 31 '16 at 17:16
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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 $\eta\le\alpha<\omega_1$.

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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.

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