Do the set S={$s \in Q: s \in [0, 1$]} have volume zero? Here, I wonder what is a good way to use the epsilon delta definition or converging sequences to show that the set S containing quotients on [0,1] have/does not have volume 0, (i.e. whether there exist a finite number of intervals which union contain all of S such that the sum of length of all intervals is less than any $\epsilon > 0$ you fix). I believe it more likely does not have volume 0 from my intuition . I am lost on where to start the proof. 
Does the idea of closure of S play a part in this proof? and how?
Also, is it possible to prove this using pigeonhole principle involving infinite rationals in one interval?
 A: Suppose $\lambda(\Bbb Q \cap [0,1]) < 1$, where by $\lambda$ I mean the volume of the set (as (sort of) defined in the comments to the question.
This would mean that we can cover $\Bbb Q \cap [0,1]$ by finitely many intervals $I_1,\ldots, I_n$ such that $\sum_{j=1}^n \lambda(I_j) < 1$ too, and we can assume these are closed intervals (as $\lambda([a,b])=\lambda((a,b))=b-a$ etc.) but then $[0,1]\setminus (\bigcup_{j=1}^n I_j)$ is a non-trivial open subset of $[0,1]$ that misses the dense set $\Bbb Q \cap [0,1]$, which is a contradiction. So $\lambda(\Bbb Q \cap [0,1])=1$. 
A: Indeed, closure does play a part, if you are going for finitely many intervals to do the covering.
The point is, if $I_1,...,I_n$ are intervals that covered $S$, say $S \subset \cup_{i=1}^n I_i$, then we can assume that the $I_i$ are closed (doesn't change their length and increases the union anyway), so that $S$ is contained in the finite union of the closed sets , which will remain closed (by finiteness of how many sets we are taking a union of). Thus, by definition of closure, we get that $[0,1]$,  the closure of $S$, is contained in the union of these intervals.
Now, by monotonicity and subadditivity of the length of intervals under taking their union, we get that the sum of the lengths of the intervals is at least $1$. Thus, under taking only finite intervals for covering, we cannot get volume zero.
If you allow infinitely many intervals, then the union is no more closed, for example, and then one checks that we can narrow the lengths as we want to make the volume zero.

Suppose $I_i$ are closed intervals such that $[0,1] \subset I_i$. We have the usual notion of length : if $I = [a,b]$ where $b \geq a$ then $l(I) = b-a$.
Now, we define the length of a finite disjoint union of intervals, by taking the sum. Thus , for example $l([0,1] \cup [2,3]) = 1+1 = 2$.
When we take the union of two intervals, either they overlap so some length is lost in the union, or the union is disjoint so retains the length. In short, it is a standard argument (extended to more intervals) to show that $l(A \cup B) \leq l(A) + l(B)$ for $A,B$ disjoint union of intervals, by checking how much each interval overlaps with another and so on.
Next, we get by induction that $l(A_1 \cup ... \cup A_n) \leq l(A_1) + ... + l(A_n)$. 
Monotonicity of length is clear : if a set contains another, it must have larger length : look just at intervals, and extend to a union of intervals again.
Now, using the definition, we get $l([0,1]) \leq l(I_1) + ... + l(I_n)$ because $[0,1]$ is contained in $\cup I_i$, so we are using both monotonicity and subadditivity here.
A: In order to provide an effective answer, just think of the volume of the complementary set, which is the whole interval but a countable union of zero-measure sets, hence the complementary set has measure one (i.e., the set you are looking for has measure zero).
