Exercise 3.5.5

This is just the complement of Exercise 3.5.4, If we had $R={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{F}_n$ then we would also have

for $G_n=F_n^c$. $G_n$ is open as it is the complement of a closed set, and since $F_n$ contains no nonempty open intervals $G_n$ is dense. This contradicts ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{G}_n\ne \varnothing$ from 3.5.4.

To be totally rigorous we still have to justify $F_n^c$ being dense. Let $a,b\in R$ with $a<b$, since $(a,b)⊈F_n$ there exists a $c\in (a,b)$ with $c\in F_n^c$ and thus $F_n^c$ is dense.