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So I am trying to figure out a proof for the following statement, but I'm not really sure how to go about it. The statement is: "Show that for every $\epsilon>0$, there exists an open set G in $\mathbb{R}$ which contains all of the rational numbers but $m(G)<\epsilon$." How can it be true that the open set G contains all of the rational numbers but has an arbitrarily small measure?

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See this answer to a related earlier question. –  Brian M. Scott Sep 23 '12 at 23:08
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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.

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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.

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You can do this for any countable set. –  ncmathsadist Sep 23 '12 at 23:42
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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....

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