If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.

One way for an irrational number $\alpha$ to be in this new set is because I threw away something like (?, $\alpha$) and ($\alpha$, ?). Each interval I removed can account for at most two numbers in this manner, and since I threw away countably many intervals, this only explains countably many irrationals that remain.

But since my [0,1] set without the cover has measure 1 - $\epsilon$, there must be more than countably many points.

So there must be another way for irrationals to remain than how I described above; but I can't think of how.

  • $\begingroup$ It will be an uncountable union of single points... $\endgroup$ – N. S. Aug 19 '17 at 3:42
  • 2
    $\begingroup$ There are many points that will remain that are not endpoints of intervals. Here is a simple example: Consider the intervals of the form $(1/(n+1),1/n)$ and $(-1/n,-1/(n+1))$ for $n=1,2,\dots$. They are all subintervals of $[-1,1]$. When you remove them, $0$ remains. But $0$ is not an endpoint of any of these intervals. Instead, it is a limit of endpoints. What you get in general is pretty complicated patterns of such limits. $\endgroup$ – Andrés E. Caicedo Aug 19 '17 at 3:51
  • $\begingroup$ That's interesting -- that is another source of numbers I hadn't thought of. That would add a another countable number of endpoints using that method, right? If you make this an answer, I would accept it -- though it would be really neat to be able to characterize many more ways such points get in. $\endgroup$ – MathStudent Aug 30 '17 at 22:23

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