Does certain open cover for rationals numbers in [0,1] cover the whole set? My question is based in the following result:
"The measure of the set of rational numbers in [0,1] is zero"
The argument asociated with this result is constructed over the fact we can list the rational numbers in a way like this:
{0} ,{1}, {1/2}, {1/3, 2/3}, {1/4  3/4}, {1/5, 2/5, 3/5, 4/5} .... and cover them (each number) with intervals with length $\epsilon/2^{n+1}$, proceed to sum them and obtains 
"Total length covered"=$\epsilon(1/2+1/4+1/8+...)=\epsilon$, Since $\epsilon$ can be made as small as we please, the measure of the rational numbers is zero. 
This is OK, but that should mean (and I want to prove that) this cover can't cover the whole [0,1], otherwise we're claiming that [0,1] has measure zero. How to prove that this cover is not enough for covering [0,1]?
 A: By the Heine-Borel theorem, if an infinite sequence of open intervals
cover the compact set $[0,1]$, then a finite subcollection do.
It is not hard to prove that if intervals $I_1,\ldots,I_n$
cover $[0,1]$ then the sum of their lengths is at least $1$.
A: It follows from Roth's theorem that for every algebraic number $\alpha \in [0,1]$ (that is, every $\alpha$ that is the root of some polynomial with rational coefficients) there is a sufficiently small $\epsilon > 0$ for which $\alpha$ won't be included in the union. That proves that $[0,1]$ is eventually not entirely covered.
One particular instance of Roth's theorem says that the inequality
$$
  \left|\alpha - \frac pq\right| < \frac1{q^3}
$$
is satisfied for only finitely many fractions $\frac pq$. The size $\frac{\epsilon}{2^{n+1}}$ that we include around the $n^{\text{th}}$ rational number $\frac{p_n}{q_n}$ decays much faster than $\frac1{q_n^3}$. So for any fixed $\epsilon$ (say, for $\epsilon = 1$) there will only be finitely many intervals containing $\alpha$.
If those intervals are around the fractions $\frac{p_1}{q_1}, \dots, \frac{p_k}{q_k}$, then for each of them we can look at the distance $\left|\alpha - \frac{p_i}{q_i}\right|$, and choose $\epsilon$ small enough so that $\alpha$ is left out of the interval around $\frac{p_i}{q_i}$. If we do this for all $k$ fractions, then we have ensured that $\alpha$ is not contained in any of the intervals.
In fact, the same argument shows that for every number with finite irrationality measure, there is a small enough $\epsilon>0$ such that it gets excluded. Only really weird transcendental numbers that have infinitely many ridiculously good rational approximations will stay in the union of this cover for all $\epsilon>0$.
