Perfect set without rationals 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).
 A: An easy example comes from the fact that a number with an infinite continued fraction expansion is irrational (and conversely). The set of all irrationals with continued fractions consisting only of 1's and 2's in any arrangement is a perfect set of irrational numbers.
A: Consider the set of reals x whose binary expansion, if you look only at the even digit places, is some fixed non-eventually-repeating pattern z. This is perfect, since we have branching at the odd digits, but they are all irrational, since z is not eventually repeating.
You can draw a picture of this set, and it looks something like the Cantor middle third set, except that you divide into four pieces, and take either first+third or second+fourth, depending on the digits of z.
Another solution: Begin with an interval having irrational endpoints, and perform the usual Cantor middle-third construction, except that at stage n, be sure to exclude the n-th rational number (with respect to some fixed enumeration), using a subinterval having irrational endpoints. By systematically excluding all rational numbers, you have the desired perfect set of irrationals.
(Hi François!)
A: It is well-known that $C$ is homeomorphic to $C \times C$, where $C$ is the Cantor set, as both are zero-dimensional compact metric spaces without isolated points. So $C$ contains uncountably many disjoint homeomorphic copies of $C$ and at most countably many of them can contain rationals...
A: 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.
A: Let $A$ be an open subset of $R$ of finite measure and containing $Q$. This is possible because $Q$ is countable. Let $B=R$ \ $A$. Now $B$ is closed, and uncountable (because it has infinite measure). Let $ C$ be the family of open real intervals that, each, have countable intersection with $B$. Then $\cup C$ is equal to $\cup D$ where $ D $ is a countable subset of $ C$, so $B$ has countable intersection with $\cup C$. The uncountable closed set $E= B$ \ $\cup C$ is perfect. Indeed, if $p \in E$ and $V$ is an open interval containing $p$, then $E \cap V$ is uncountable. 
