# Complement of a countable open cover of the rationals

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.

• 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 – Ishfaaq Jun 2 '14 at 1:43
• @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. – Cheerful Parsnip Jun 2 '14 at 1:46
• The seeming paradox you've noticed is related to the fact that the rationals have measure $0$. – Cheerful Parsnip Jun 2 '14 at 1:47
• 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? – coolpapa Jun 2 '14 at 1:51
• @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. – Michael Hardy Jun 2 '14 at 2:27

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 $q$.
• 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. – Cheerful Parsnip Jun 2 '14 at 1:51