# 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]?

• Well, you just did. What more more needs to be proven? Such a cover can have arbitrarily small measure but the measure of $[0,1] = 1$ is not arbitrarily small. So it can't cover $[0,1]$. There is nothing more you need to prove. You are COMPLETELY done! – fleablood Apr 15 '18 at 7:31
• You're right. I just wanted to see if there is a way of characterizing the numbers I would exclude of this covering. – Mauro Cruz Apr 15 '18 at 7:42

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 $\epsilon>0$.

• I guess when you are refering to algebraic numbers, you're talking about the irrational ones. On the other hand, do you mean that those irrational numbers covered for a number INFINITE of intervals never "escape" from this cover? – Mauro Cruz Apr 15 '18 at 7:26
• Right, the irrational ones. Numbers covered by an infinite number of intervals might escape from this cover, but it's harder to prove: if we try to choose an $\epsilon$ small enough for each of the infinitely many intervals, we might end up having to choose $\epsilon=0$. – Misha Lavrov Apr 15 '18 at 16:17

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 $1$.

• I agree, but I'm looking for how to show this particular cover does not cover [0,1] – Mauro Cruz Apr 15 '18 at 7:04
• @MauroCruz This is a proof by contradiction. If it did cover $[0,1]$, then the sum of its lengths would be at least $1$, so once the sum of lengths is less than $1$, it cannot cover $[0,1]$. – Misha Lavrov Apr 15 '18 at 7:12