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.
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.)