Find an open set which has finite length and is super set of rational numbers We define the length of the open set as the difference between the ends.
$l[(a,b)]=b-a$ and
$l[(a,b)\cup(c,d)]\leq l[(a,b)]+l[(c,d)]$
We have to find one open set $U$ such that $l(U)<\infty$ and $Q\subset U$
 A: You're probably running into problems because of too much detail. In the usual (AFAIK) approach, the only relevant property of $\mathbb{Q}$ is that it is countable; remembering how $\mathbb{Q}$ is topologically embedded into $\mathbb{R}$ is an irrelevant detail that only gets in the way of understanding.
The usual method proves:
Theorem: Let $S$ be a countably infinite subset of $\mathbb{R}$. Then there is an open set of finite length containing $S$.
In fact, you can impose an upper bound; for any $\epsilon > 0$ you can find a set of length less than $\epsilon$.
The construction is basically figure:


*

*Figure out how to do this for $\mathbb{N}$

*Rearrange the intervals so that they cover of $S$ instead of $\mathbb{N}$


*

*This uses the existence of a bijection between $S$ and $\mathbb{N}$

*Recognize it's fine to make the rearranged intervals overlap; that just decreases the total length of the ensuing set.


A: Let {$q_n$} be an enumeration of the rationals.
$\cup$ { $B(q_n,1/2^n) \ \ \ \ $ : n in N }
is an open set containing the rationals that has a measure ("length") <= 2 as determined by adding the lengths of all the intervals.
