Fall 2001 Problem 6

Exercise 12.4 (Bounded only on the irrationals)   Show that there does not exist a sequence of continuous functions $f_n : \mathbb{R} \to \mathbb{C}$ such that the sequence $\{f_n(x)\}$ is bounded if and only if $x$ is irrational. (Hint: Show that the set $\{x \;\vert\;$   ${f_n(x)}$ is bounded$\}$ is an $F_\sigma$.)

Proof. Per the hint, we write

$$\{x \;\vert\; \{f_n(x)\} \} = \bigcup_{M=1}^\infty\bigcap_{n=1}^\infty \{x \;\vert\; \vert f_n(x)\vert \leq M\}$$ (12.38)

The intersection is over closed sets, so it is closed, indicating this is indeed an $F_\sigma$ set. At this point we are done because $\mathbb{R}\setminus\mathbb{Q}$ is definitely not an $F_\sigma$. In fact, $\mathbb{Q}$ is an $F_\sigma$ set, because it is a countable union of points, so that the irrationals form a $G_\delta$ set. $\qedsymbol$