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How can I show that there are uncountably many intervals in R?

I think contradiction would work by assuming there are countably many intervals and using that there are countable number of rational number in R, but I'm not sure how to proceed.

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  • $\begingroup$ Did you attempt to exhibit an uncountable family of intervals? Shouldn't be hard. $\endgroup$ – anon Jul 18 '13 at 2:41
  • $\begingroup$ $\Bbb{R}$ contains only countably many open intervals. $\endgroup$ – Hanul Jeon Jul 18 '13 at 2:44
  • $\begingroup$ Presumably you can use the fact that there are uncountably many reals. You should be able to find a function $I(x)$ such that $I(x)$ is always an interval, and $x\ne y$ implies $I(x)\ne I(y)$. Remark: Contradiction is sometimes overrated. $\endgroup$ – André Nicolas Jul 18 '13 at 2:56
  • $\begingroup$ ^ And potentially the domain of $I(x)$ need not even have domain all of $\bf R$. @tetori: No, there are both uncountably many open intervals and uncountably many closed intervals. $\endgroup$ – anon Jul 18 '13 at 2:57
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    $\begingroup$ @anon Oh, my mistake. There is no uncountable collection of pairwise disjoint open intervals in $\Bbb{R}$. $\endgroup$ – Hanul Jeon Jul 18 '13 at 3:01
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Recall that a set is uncountable iff it has an uncountable subset. From the set of all possible intervals in $\Bbb{R}$, we consider the subset of such intervals of the following form: $$ S=\{(x,\infty) \mid x\in \Bbb{R}\} $$ It suffices to show that $S$ is uncountable. To do this, we construct a bijection $f\colon S\to\Bbb{R}$ defined by: $$ f((x,\infty))=x $$ Hence, since $\Bbb{R}$ is uncountable, so too is $S$, as desired.

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  • $\begingroup$ @anon I've changed my argument entirely. Hopefully this works now. $\endgroup$ – Adriano Jul 18 '13 at 3:05
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Consider $(\alpha ,\beta)\in \mathbb{Q}^c\times\mathbb{Q}^c $. Each of these pair will give you distinct intervals. Hence set of all intervals contains an uncountable set. Therefore it is itself uncountable.

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