# An example of an open set of real numbers containing the set of rational and having complementary non-countable.

An example of an open set of real numbers containing the set of rational and having complementary non-countable.

I can not think of any set that satisfies this condition.

• Se here for a previous answer – Old John Sep 19 '13 at 20:42

Fix an enumeration $\{r_k\}_{k \in \Bbb{N}}$ of $\Bbb{Q}$ and let $O_k = (r_k - \frac{1}{2^k}, r_k + \frac{1}{2^k})$. Consider $O = \bigcup_k O_k$. Do you see why $O$ contains every rational number, while also having very large complement?
• sorry, but I can not see that the complement of $O$ is great. – Henfe Sep 19 '13 at 20:47
• @Henfe Each interval $O_k$ has length $1/2^{k -1}$. Adding up all the lengths gives a finite number (in fact, \$2@), since it's a geometric series. – user61527 Sep 19 '13 at 20:48