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Prove that the set of all open intervals with rational endpoints are countable, I do not know exactly what shall I do, Could anyone help me please?

Emptymind
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    This cardinality will be less than or equal to the cardinality of $\mathbb{Q} \times \mathbb{Q}$ and greater than or equal to the cardinality of $\mathbb{Q}$. So it suffices to show that $| \mathbb{Q} \times \mathbb{Q}| = | \mathbb{Q}|$. – Kaj Hansen Mar 28 '17 at 19:17
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    All comments and answers from this other recent question What will be $ | \mathbb Q^2|$? apply. – dxiv Mar 28 '17 at 19:19
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    Then interval (a,b) is uniquely determined by the ordered pair <a,b> where a < b. the set of all <a,b> were b>a and a,b are rational is a subset of Q x Q. which is countable. – fleablood Mar 28 '17 at 19:36

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The set of rational numbers $\mathbb{Q}$ is countable. Furthermore, you should know that a finite product of countable sets is countable, so $\mathbb{Q}\times\mathbb{Q}$ is countable. (If you don't know this, prove it!) You can construct a bijection from your set to $\mathbb{Q}\times\mathbb{Q}$ the natural way, so that set will also be countable.

Kevin Long
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  • you that the bijection is constructed in the natural way? – Emptymind Apr 07 '17 at 14:42
  • By natural, I just mean the easiest way to do it? In this case, as fleablood described in the comments, you map an open interval $(a,b)$ with rational endpoints to the ordered pair $(a,b)\in\mathbb{Q}\times\mathbb{Q}$. You can show fairly easily that this is an injection, so $|\mathbb{Q}\times\mathbb{Q}|\geq |I|$ where $I$ is the set of open intervals with rational endpoints. (This isn't a bijection, since if $a>b, (a,b)$ is not an interval (or, at best, it's a degenerate interval). – Kevin Long Apr 07 '17 at 15:17