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I came across the following problem on open sets:

Let $\mathcal{C}$ be a set of nonempty open sets in $\mathbb{R}$ such that for $U,V \in \mathcal{C}$, either $U=V$ or $U \cap V = \emptyset$. Prove that either $\mathcal{C}$ is finite or the sets in $\mathcal{C}$ can be listed in a sequence $(U_n)$.

Now we can enumerate the rational numbers using Farey sequences. So choose a rational number from each $U \in \mathcal{C}$. Form a neighborhood about each rational number from the open sets and see if they (the neighborhoods) intersect or not.

So it suffices that if we choose a rational number from each open set, by countability of the rationals $\mathcal{C}$ is countable?

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What is your question? Seems unclear to me. –  Sunni Jun 27 '11 at 15:13
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Damien: a little piece of advice. The number of questions you can ask here is limited by the software to 50 a month, see here. You asked 26 questions in about a week, so maybe you want to slow down a little. –  t.b. Jun 27 '11 at 15:15
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The "Form a neighborhood $\dots$ or not." sentence is superfluous. The way one enumerates the rationals doesn't make any difference to the argument. But Farey sequences don't enumerate the rationals. –  André Nicolas Jun 27 '11 at 16:10
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1 Answer

Yes, that works. You seem uncertain, but you explicitly show an injection into a countable set.

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