Exercise 2.18

Yes. Here is one example: Choose two irrational numbers, e.g. $\alpha =\sqrt{2}$ and $\beta =\sqrt{3}$, and let $\{a_n\}$ be an enumeration of the rational numbers in $E_1=[\alpha ,\beta ]$.

Let $r_1$ be the first element of $\{a_n\}$ that is in $E_1$, and pick irrational numbers ${\alpha }_1<r_1$, ${\beta }_1>r_1$ such that $\alpha <{\alpha }_1<r_1<{\beta }_1<\beta$. By removing the interval $({\alpha }_1,{\beta }_1)$, we end up with a union of closed intervals

By repeating this process for each closed interval in this union and continuing this process in the same vein as the construction of the Cantor set, we end up with a sequence of compact sets

Now consider $E={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n$. The reasoning on p. 41 carries through with small modifications, so that we can deduce that $E$ is perfect. By construction, there are no rational numbers in $E$.