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.

  • $\begingroup$ Se here for a previous answer $\endgroup$ – 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?

  • $\begingroup$ sorry, but I can not see that the complement of $O$ is great. $\endgroup$ – Henfe Sep 19 '13 at 20:47
  • $\begingroup$ @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. $\endgroup$ – user61527 Sep 19 '13 at 20:48

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