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All rational numbers of the unit interval [0, 1] can be covered by countably many intervals, such that the $n$-th rational is covered by an interval of measure $1/10^n$. There remain countably many complementary intervals of measure $8/9$ in total.

Does each of the complementary intervals contain only one irrational number? Then there would be only countably many which could be covered by another set of countably many intervals of measure $1/9$.

Is there at least one of the complementary intervals countaining more than one irrational number? Then there are at least two irrational numbers without a rational between them. That is mathematically impossible.

My question: Can this contradiction be formalized in ZFC?

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    I'm not sure about your question, but in any interval there are infinitely many irrational numbers...2012-04-15
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    I wouldn't be so sure about "countably many complementary intervals". Imagine removing just the rationals from $[0,1]$. This way you remove only countably many points. The set of irrationals that remains, contains no non-degenerate intervals, i.e. its connected components are singletons. An uncountable number of singletons. So, I'm pretty sure the construction you mention does something similar.2012-04-15
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    I remember [this](http://mathoverflow.net/questions/88267/covering-the-rationals-a-paradox) which seems to be somewhat related to this question.2012-04-15
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    Thanks Asaf for the hint. But it is obvious that the answer given there with "indeed" and "ín fact" is not mathematics. By induction it can be proven that n intervals yield n-1 complementary intervals. This is true for every n and, therefore, for all n. We count the complementary intervals. Otherwise, if induction would not reach till the first infinity, Cantor's counting of the rationals could not cover all rationals; it woulde be invalid too - like the counting of the complementary intervals here. If bijection is denied in one example then it can also be denied in the other one.2012-04-15
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    But you only prove that by induction for *finite* $n$. Your claim is about an infinite collection of intervals. **Induction does not prove the infinite case!**2012-04-15
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    Much the same posted to MO, http://mathoverflow.net/questions/94105/why-is-counting-remaining-intervals-different-from-counting-rationals2012-04-15
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    @ Gery: Asaf hinted me to the recent discussion. The answer there was poor and unmathematical. Indeed and in fact ma be used in real life, but not in mathematics. @ Asaf: The pairing function or bijection applied for all finite cases proves, according to set theory, countability. Therefore there are countably many complementary intervals.2012-04-15
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    Who is Gery? What is ma?2012-04-15
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    @BelsaZarkin, you are wrong. There is **no interval left behind**. Every interval contains at least **one** rational number, so if you remove all rational numbers (let alone a bunch of intervals containing all of them), there can be **no** interval left over.2012-04-20

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