Let $a,b\in\mathbb{R},$ and let $a<b.$ Are there uncountably many pairwise disjoint open intervals of $I=(a,b)?$ I know that the answer is no since if there were uncountably many pairwise disjoint open intervals of $I,$ then, from the density of $\mathbb{Q}$ in $\mathbb{R},$ we would get that each of these intervals would have atleast one rational in them implying that $\mathbb{Q}$ is uncountable. I am looking for a different proof, though.
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3Why would you need a different proof? – Arthur May 15 '23 at 09:55
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1@Arthur, I just want a different way of seeing this. – aqualubix May 15 '23 at 09:57
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The answer is obviously "yes": take the singletons. You are probably meaning "non-trivial subintervals", not just "subsets". – Anne Bauval May 15 '23 at 09:57
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1Are you asking about subsets or open subsets? – Brian Moehring May 15 '23 at 09:57
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@AnneBauval, edited, please check. – aqualubix May 15 '23 at 09:58
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@BrianMoehring, I'm asking about intervals (I edited the question). – aqualubix May 15 '23 at 09:59
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1"Intervals" is not sufficient. Singletons are (closed) intervals. – Anne Bauval May 15 '23 at 09:59
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@AnneBauval, open intervals, then. Sorry, I'm new to analysis. – aqualubix May 15 '23 at 10:00
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This very FAQ always has the same answer: the one you don't want. – Anne Bauval May 15 '23 at 10:10
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@AnneBauval, then I guess I'm stuck with this proof for now. I'll delete the question. – aqualubix May 15 '23 at 10:13
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1Does this answer your question? If $\sum_{i\in F} a_i$ is finite then $a_i =0$ except for countably many $i\in F$ (apply it on the lengths of your intervals) – Anne Bauval May 15 '23 at 10:13