The sum of the lengths of open intervals in a finite covering of $\mathbb{Q}\cap [0,1]$ is $\geq 1$ Here is what I have so far:
Arrange the intervals $I_k = (a_k,b_k)$ such that $0\in (a_1,b_1), 1\in (a_n,b_n)$. Then, since the rationals are dense, $a_{k+1} \leq b_k$ for all $k$. 
I saw another solution that said:
Replacing $I_k$ by smaller intervals if need be, we may assume that $a_{k+1} = b_k$ for all $k$. Then
$$
\sum_{i=1}^n l(I_k)
$$
becomes a telescoping sum equal to $b_n - a_1 \geq 1$, 
However, I don't see why it is ok to just replace the intervals with smaller ones.  
Could someone please explain?
 A: Prove that:
The sum of lengths of finite many open intervals covering $\mathbb Q\cap[a,b]$ is at least $b-a$.
We use induction on the number $n$ of open intervals in the cover. 
For $n=1$ is obvious. Suppose it is true for all $k<n$ and the intervals $I_1,\ldots, I_n$, cover $[a,b]$. If $I_1,\ldots, I_{n-1}$ also cover $[a,b]$, we are done. If not, then the intervals $I_1,\ldots, I_{n-1}$ cover a set of the form
$$
[a,b_1]\cup [a_1,b]
$$
in which case the inductive hypothesis implies that:
$$
\sum_{k=1}^{n-1} \ell(I_k)\ge (b-a_1)+(b_1-a), \tag{1}
$$
and also that 
$$
(b_1,a_1)\subset I_n \tag{2}
$$
Combination of (1) and (2) provides that
$$
\sum_{k=1}^{n} \ell(I_k)\ge (b-a).
$$
A: Since $a_{k+1} \le a_k$, the length of $(a_k,b_k) \cup (a_{k+1},b_{k+1})$ is equal to the length of $(a_k,a_{k+1}) \cup (a_{k+1},b_{k+1})$. Hence, you can replace $I_k=(a_k,b_k)$ with smaller interval $(a_k,a_{k+1})$. Repeat the process finite number of times until you have $n$ disjoint open intervals.
A: When we want to prove a sum is greater than some number, we are free to replace terms in the sum with smaller ones.  The new sum will be smaller than the original one, but if it is greater than the target we are OK.  If we just take a case with two intervals, we might have the original two intervals be $[0,0.6], [0.4,1]$  The sum of the lengths is $1.2$.  The proof says that if the intervals cover $[0,1] \cap \Bbb Q$, so will the set of intervals $[0,0.6],[0.6,1]$.  We have thrown away the overlap.  We know how to add up the intervals that just meet at the endpoints, for which we get $1$, so the sum of the lengths of the original intervals is at least that high.
