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Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals.

Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrational. For instance, if s = sqrt(2) then we might have the two sub-intervals ( s-I/2, s ) and ( s, s+I/4 ) and s would be uncovered.

But if we have a countable number of open sub-intervals with at most one (irrational) number between each sub-interval, that would mean we've covered all of the real line except for the complement, a countable number of irrationals, with an interval of length 1, which is absurd.

So there's a flaw in the reasoning but I don't see exactly where. Where is it?

Thanks.

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  • $\begingroup$ If what you call absurd is the countable nature of the irrationals in $I$, note that every such open interval which cover the rationals contain an extraordinary number of irrationals in them. You cannot have an open interval in $\Bbb R$ which does not contain irrationals. So you have not counted them $\endgroup$
    – Ishfaaq
    Jun 2, 2014 at 1:43
  • $\begingroup$ @Ishfaaq: you seem to be misunderstanding the question. I have to admit, I needed to read it several times before I figured out what the OP was getting at. $\endgroup$
    – Cheerful Parsnip
    Jun 2, 2014 at 1:46
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    $\begingroup$ The seeming paradox you've noticed is related to the fact that the rationals have measure $0$. $\endgroup$
    – Cheerful Parsnip
    Jun 2, 2014 at 1:47
  • $\begingroup$ Your notation also makes your question hard to understand. What does $I/2$ mean? Is it an interval? If so, what does $s - I/2$ mean? $\endgroup$
    – coolpapa
    Jun 2, 2014 at 1:51
  • $\begingroup$ @ArtD You're using terms rather clumsily. You say "divide it into countable subintervals". It appears that you probably mean countably many subintervals. The term "countable subintervals" litearally means "subintervals, each one of which is countable". And the question of what "$I/4$", etc., means has been rasied in other comments. It appears to me that you probably mean you want to cover the rationals with a set of intervals the sum of whose lengths is finite. That can be made explicit. $\endgroup$
    – Michael Hardy
    Jun 2, 2014 at 2:27

1 Answer 1

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The intervals must be very overlapping. If $q$ is covered by an interval of length $2x$, there is a rational $q'\in(q-x,q+x)$. And the interval covering $q'$ must have intersected with the one covering $q$.

But even if you did manage to make this cover without overlapping intervals (e.g. at each step pick the smallest rational which you can cover without overlaps), your mistake is that you falsely conclude that the intervals must be separated by at most a single irrational number, and that the complement is just these irrational numbers. Most of the irrational numbers are not endpoints of these intervals, but rather limit points of them.

(The same thing happens with the Cantor set, most points of the Cantor set are not interval endpoints from those that we remove, but rather limit points of these endpoints.)

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  • $\begingroup$ I interpreted the question to mean that the paradox was that you can cover "most" of the real line with a union of intervals of length $1$. But then I guess your question addresses the OP's supposition that only countably many points are omitted, when in actual fact uncountably many are. $\endgroup$
    – Cheerful Parsnip
    Jun 2, 2014 at 1:51
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    $\begingroup$ I think it's a bit of both. I admit that like you, I had to read this question a couple of times before writing my answer. When I first read it, I thought it was a duplicate of an older answer of mine. But then I saw that it's not. (Or at least not the one I had in mind...) $\endgroup$
    – Asaf Karagila ♦
    Jun 2, 2014 at 1:59

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