Open Measurable Sets Containing All Rational Numbers

Solution 1:

To elaborate on other answers, let $\{r_n\}_{n \in \mathbb{N}}$ be an enumeration of all rational numbers. This is possible because $\mathbb{Q}$ is countable. For each number $r_n$, pick the open interval $(r_n - 2^{-n-1} \epsilon, r_n + 2^{-n-1} \epsilon)$. Each open interval is an open set. The union of all such intervals is also open. The measure of each interval is its length $2^{-n} \epsilon$. If $G$ is the union of these intervals, we have:

$$ m(G) \le \sum_{n=1}^{\infty} 2^{-n} \epsilon = \epsilon $$

Where the sum above is a geometric series.

Solution 2:

The rationals are denumerable. Enumerate them and then choose an interval of length less than $\epsilon/2^n$ for $n\ge 1$. Let $G$ be the union of these intervals. It has measure less than $\epsilon$ and contains all rationals.

Solution 3:

Hint: if you order the rationals, you can put an interval around each successive one and take the union for your set. If the intervals decrease in length quickly enough....