Disjoint open intervals covering the rationals In this answer and this question, we saw how to construct open intervals covering $\Bbb Q \cap [0,1]$ with total length being $\epsilon$, an arbitrarily small positive real number.
If we do it with disjoint open intervals, can we still make the total length arbitrary?
In symbols:
$$\forall \epsilon > 0: \exists \langle I_n \rangle: \left(\left(\sum_{i \in I_n}|i| < \epsilon \right) \land \left(\forall a,b \in \Bbb N: I_a \cap I_b = \varnothing \right) \land \left(\Bbb Q \cap [0,1] \subset \bigcup I_n\right)\right)$$
 A: You can make it arbitrarily small even with disjoint open intervals.
Take an arbitrary enumeration of $\mathbb Q\cap [0,1]$, as $\{q_i\}_{i=0}^\infty$. We will make sure that each interval we take has irrational valued endpoints. Also take $\gamma=\epsilon/2$
At step $i$ we have a finite number of open intervals that have irrational endpoints call them $Q_i=\{(a_j,b_j)\;|\; (j<i)\;\wedge\; (\forall j<i \exists l<i \;q_j\in(a_l,b_l))\}$ which are all disjoint and the sum of whose sizes add up to some $\epsilon_i<\gamma$. We also have $Q_i\subseteq Q_k$ when $i<k$.
We now need to create $Q_{i+1}$ which still satisfies all the criteria. 
If $q_{i+1}$ is in one of the intervals that is in $Q_i$ we set $Q_{i+1}=Q_i$ and are done.
If $q_{i+1}$ is in none of the intervals in $Q_i$ we look at the distances $d_j$ of $q_{i+1}$ from $(a_j,b_j)$ for all $j<i+1$. There are only finitely many such distances and since $a_j,b_j$ are all irrational none of those distances is 0. Take the smallest of those distances and call it $D_i$. Let $s=\min\{D_i,(\gamma-\epsilon_i)/2\}$. Than let us take some $s^\prime<s/2$ irrational and define $a_i=q_i-s^\prime$ and $b_i=q_i+s^\prime$. Define $Q_{i+1}=Q_{i}\cup \{(a_i,b_i)\}$. 
Note that the fact that $s^\prime<s/2$ makes sure that $|(a_i,b_i)|<s$ and so $\epsilon_{i+1}=\epsilon_i+|(a_i,b_i)|<\gamma$. At the same time the same condition ensures that $(a_i,b_i)\cup (a_j,b_j)=\emptyset$ for all $j<i$. Obviously $q_i\in(a_i,b_i)$.
Now we just define $Q=\bigcup_{i=0}^\infty Q_i$.
Note
This only shows you can make the total length arbitrarily small. Not arbitrary. I'm not sure how to hit any specific $\epsilon$.
