Sequence of unbounded continuous functions

Question

I'm having trouble with an exercice from Rudin's Real and Complex Analysis:

Does there exist a sequence $(f_n)$ of continuous positive functions on $\Bbb R$ such that $f_n(x)$ is unbounded if and only if $x \in\Bbb Q$ ?

What if $x \in\Bbb R \setminus\Bbb Q$?

Now, replace unbounded by $f_n(x) \rightarrow \infty $ as $ n \rightarrow \infty$ and answer the analogues to the previous questions.

I don't know very well how to construct such functions, as they would have to be continuous. My guess is that you can do it for $x \in\Bbb Q$ , as their measure is zero, but I don't know how to prove that you can or cannot do it.

Edit:

I have found the answers for $f_n(x) \rightarrow \infty $, but I don't understand yet how to prove the first cases. I think I have to use Baire's theorem but I don't know how to get there.

Answer 1

In the first case such a sequence can't exist. Note that $\mathbb{Q}$ is not a $G_\delta$ set, since if $\mathbb{Q} = \cap_n V_n$ with $V_n$ open, then $\mathbb{R} = \{ r_m \} \cup (\cup_n V_n^c)$ would be of first category (this is discussed more detailed here). On the other hand, the set $\cap_m \cup_n \{x : f_n(x) > m \}$ of points at which a sequence of positive continuous functions is unbounded is a $G_\delta$ set.

In the second case there is such a sequence. Let $\mathbb{Q} = \{q_k\}$ and consider

$$f_n(x) = \min_{1 \leq k \leq n} \{k + n \vert x - q_k \vert \} \geq 1.$$

Then for each $q_m \in \mathbb{Q}$ and $n \geq m$, one has $f_n(q_m) \leq m + n \vert q_m - q_m \vert = m$, hence $\{f_n(q_m)\}$ is bounded. On the other hand, if $x \in \mathbb{Q}^c$, then given $M> 0$, there is some $N = N(M) > M$ such that $n \vert x - q_i \vert > M$ for all $1 \leq i \leq M$ provided $n > N$. Therefore,

$$f_n(x) = \min_{1 \leq k \leq n} \{k + n \vert x - q_k \vert \} > M,$$

that is, $f_n(x) \longrightarrow \infty$.