Perfect set in $\mathbb{R}$ which contains no rational number 
Possible Duplicate:
Perfect set without rationals 

Does there exist a nonempty perfect set in $\mathbb{R}$ which contains no rational number?
This problem is on p.44 PMA - Rudin
I found a proof of this on google but the proof is not 'suitable' for me, since the proof uses the concept of 'measure', which is in chapter 11, while this problem is on chapter 2.
Can one show this by direct construction?
 A: Since the rational are countable, let $a_n$ enumerate all the rationals. Let $B_{2^{-n}}(a_n)$ be the open interval centered at $a_n$ of radius $2^{-n}$. Let $A = \bigcup_{n \in \mathbb{N}} B_{2^{-n}}(a_n)$. $A$ is open since it is it is a union of open sets. Note that $\mathbb{Q} \subset A$.  Clearly $A$ is not all of $\mathbb{R}$ because "length" of $A$ is less than or equal to $2\sum_{n = 1}^\infty 2^{-n} < \infty$. (This is essentially the measure idea.) Because of this and fact that $\mathbb{R}$ is uncountable, you have that $C:= \mathbb{R} - A$ is an uncountable set. $C$ closed since it is the complement of an open set. Also $C \cap \mathbb{Q} = \emptyset$ since $\mathbb{Q} \subset A$. 
Finally, by the Cantor Bendixson Theorem (Exercise 28 on page 45) which states that every uncountable closed set is (uniquely) the union of a perfec set and a countable. Hence there exists a perfect set $C'$ and a countable set $F$ such that $C = C' \cup F$. Since $C \cap \mathbb{Q} = \emptyset$, you also have $C' \cap \mathbb{Q} = \emptyset$. Hence $C'$ is a nonempty perfect set that does not contain any rational numbers.
By the way $C'$ can be constructed, look at the proof of the Cantor Bendixson Theorem to see exactly how it is made. 
