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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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closed as too localized by Zev Chonoles Jul 8 '12 at 14:38

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I'm not sure about your question, but in any interval there are infinitely many irrational numbers... –  Asaf Karagila Apr 15 '12 at 10:31
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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. –  Dejan Govc Apr 15 '12 at 10:59
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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! –  Asaf Karagila Apr 15 '12 at 11:35
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Much the same posted to MO, mathoverflow.net/questions/94105/… –  Gerry Myerson Apr 15 '12 at 12:48
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Who is Gery? What is ma? –  Gerry Myerson Apr 15 '12 at 23:33

1 Answer 1

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)

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@Belsa No. The definition of a totally disconnected space is a space in which every connected space is a singleton. E.g. a discrete space. Are you saying that every discrete space is countable? –  Asaf Karagila Apr 15 '12 at 12:52
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@Belsa: Yes. The intersection of countably many intervals may result in a totally disconnected space, e.g. the Cantor set. –  Asaf Karagila Apr 15 '12 at 18:04
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@Belsa: I hope that you are willing to agree with me on the following: (1) Every rational number is inside the open cover. (2) Every number not in the open cover is irrational. (3) In any interval which contains more than one point there is a rational number. (4) If the set of the points not inside the open cover does not contain rational numbers then there is no interval $(a,b)$ which is a subset of that set. (5) Every countable set is of measure zero. (6) If we removed points of measure $\le\frac19$ then we are left with uncountably many points. (cont...) –  Asaf Karagila Apr 15 '12 at 18:18
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@BelsaZarkin, I heartily second Asaf's last claims. I read through this entertaining discussion and I everything Asaf said was correct, whereas most of what you said either meant nothing, or was blatantly false. I would recommend being a little more modest, and actually trying to understand what Asaf had to say. There is not a single interval contained in the complement of your open set. –  Bruno Joyal Apr 20 '12 at 7:04
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Fairly soon, there will be an uncountable infinity of comments on this answer, and then we'll be able to test all the theories directly. –  Gerry Myerson Apr 20 '12 at 13:07

protected by Asaf Karagila May 28 '12 at 10:03

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