We will construct a nonempty perfect set contained in $\mathbb{R}$ that contains no rational number.

We will begin with a closed interval, and then, imitating the construction of Cantor set, we will inductively delete each rational number in it together with an open interval. We will do it in such a way that the end points of the open intervals will never be deleted afterwards.

Let $E_0 = [b_0, a_0]$ for some irrational numbers $a_0$ and $b_0$ , with $b_0<a_0$ . Let $\{ q_1, q_2, q_3, \ldots \}$ be an enumeration of the rational numbers in $[b_0, a_0]$ . For each $q_i$ , we will define an open interval $(a_i,b_i)$ and delete it.

Let $a_1$ and $b_1$ be two irrational numbers such that $b_0<a_1<q_1<b_1<a_0$ . Define $E_1 = E_0\backslash (a_1,b_1)$ .

Having defined $ E_1,E_2,\ldots,E_n$ , $ a_1,a_2,\ldots,a_n $ and $ b_1,b_2,\ldots,b_n $ , let's define $a_{n+1}$ and $b_{n+1}$ :

If $\disp q_{n+1}\in \bigcup_{k=1}^n (a_k,b_k)$ then there exists an $i\leq n$ such that $q_{n+1}\in (a_i,b_i)$ . Let $a_{n+1}=a_i$ and $b_{n+1}=b_i$ .

Otherwise let $a_{n+1}$ and $b_{n+1}$ be two irrational numbers such that $b_0<a_{n+1}<q_{n+1}<b_{n+1}<a_0$ , and which satisfy: $$ \disp q_{n+1} - a_{n+1} < \min_{i=0,1,2,\ldots,n}\{|q_{n+1} - b_i|\}$$ and $$\disp b_{n+1} - q_{n+1} < \min_{i=0,1,2,\ldots,n}\{|a_i - q_{n+1}|\}. $$

Now define $E_{n+1}=E_n\backslash (a_{n+1},b_{n+1})$ . Note that by our choice of $a_{n+1}$ and $b_{n+1}$ any of the previous end points are not removed from $E_n$ .

Let $\disp E = \bigcap_{n=1}^\infty E_n$ . $E$ is clearly nonempty, does not contain any rational number, and also it is compact, being an intersection of compact sets.

Now let us see that $E$ does not have any isolated points. Let $x\in E$ , and $\epsilon>0$ be given. If $x\neq a_j$ for any $j \in \{0,1,2,\ldots\}$ , choose a rational number $q_k$ such that $x<q_k<x+\epsilon$ . Then $q_k\in(a_k,b_k)$ and since $x\in E$ we must have $x<a_k$ , which means $a_k \in (x, x+\epsilon)$ . Since we know that $a_k\in E $ , this shows that $x$ is a limit point. Otherwise, if $x = a_j$ for some $j$ , then choose a $q_k$ such that $x-\epsilon < q_k < x$ . Similarly, $q_k \in (a_k, b_k)$ and it follows that $b_k\in (x-\epsilon, x)$ . We have shown that any point of $E$ is a limit point, hence $E$ is perfect.