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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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    @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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There is a mistake in your argument, which follows from the fact that the complement of this union contains no interval.

Let $C=\bigcup I_n$ the open set which covers the rationals whose measure is at most $\frac19$. You are considering $B=[0,1]\setminus C$, which is a subset of the irrationals.

If $B$ would contain an open interval then it would contain a rational number. Since there are no rational numbers in $B$ there is no interval subset. The irrational numbers form a totally disconnected space, namely every connected component is a singleton.

Note that by DeMorgan we have $B=\bigcap [0,1]\setminus I_n$, this is an intersection of closed sets. Indeed for every $n$ we have that $[0,1]\setminus (I_1\cup\ldots\cup I_n)$ contains an interval, in which there are infinitely many rational and irrational numbers.

The limit, however, is not require to have the properties of the sequence, and we have that $B$ contains no interval.

(This thread can be useful here: Fake induction proof)