TITLE: Perfect set without rationals
QUESTION [35 upvotes]: Give an example of a perfect set in $\mathbb R^n$ that does not contain any of the rationals.
(Or prove that it does not exist).

REPLY [13 votes]: Just consider a translation of Cantor set $C$, denote as $E=C+\{x_0\}$. The perfectness of $E$ is trivial due to the perfectness of $C$. To make $E\cap\mathbb{Q}=\varnothing$, we need to choose an $x_0\notin \mathbb{Q}-C$. The only thing left is to show $\mathbb{Q}-C\neq\mathbb{R}$, i.e. $\mathbb{Q}+C\neq\mathbb{R}$. By Baire Category theorem
$$\mathbb{Q}+C=\bigcup_{r\in\mathbb{Q}}\{r\}+C$$
can't have any interior point, since $\{r\}+C$ don't have any interior point, for any $r\in\mathbb{Q}$. The conclusion follows.