On the existence of a sequence of positive continuous functions Does there exist a sequence $(f_n)$ of positive continuous functions on $\mathbb{R}$ such that
$f_n(x) \rightarrow \infty$ if and only if $x \in \mathbb{Q}$?
If $f_n(x) \rightarrow \infty$ is replaced by $f_n(x)$ is unbounded, then the answer is no. This follows from Baire's theorem.
Thank you,
 A: See the paper, First Class Functions, in the American Mathematical Monthly 98 (March, 1991) 237-240. 
EDIT: Let $f_n(x)=n$ if $x=p/q$ with $q\le n$, $f_n(x)=0$ if $x=(p/q)\pm n^{-4}$ with $q\le n$, and let $f_n$ be piecewise linear between these points. So $f_n$ is continuous, mostly zero, but with a sharp spike at each rational. Clearly $f_n(x)$ goes to infinity with $n$ at all rational $x$. If $x$ is irrational and has only finitely many rational approximations $p/q$ such that $|x-(p/q)|\le q^{-4}$ (and this is all $x$ save a set of measure zero), then $f_n(x)=0$ for all $n$ sufficiently large. If $x$ has infinitely many rational approximations with $|x-(p/q)|\le q^{-4}$, then $f_n(x)=0$ for most $n$ (those that are far from a $q$ which gives a good approximation, and those $q$ are guaranteed to be few and far between), but is occasionally quite large, so $f_n(x)$ has no limit, finite or infinite. 
A: An explicitely wrong solution is as follows: Choose a bijection between $\mathbb N$ and $\mathbb Q$ and denote by 
$\mathbb Q_n$ the image of $\{1,\dots,n\}$ under this bijection. Choose also a sequence $s_n(x)$ of continuous
functions on $\mathbb R_{\geq 0}$ with limit $\lim_{n\rightarrow\infty}s_n(x)=\infty$
if $x=0$ and with limit $0$ otherwise and consider the sequence of continuous functions $s_n(d(x,\mathbb Q_n))$
where $d(x,\mathbb Q_n)$ denotes the distance of $x$ to the finite set $\mathbb Q_n$
corresponding to the first $n$ rational numbers under the choosen bijection.
The limit is then clearly $\infty$ on rationals. However Malik Younsi objected correctly
that one can say nothing on the limit for irrationals and Myerson's Monthly paper 
states that the limit on irrationals cannot be bounded for all irrationals.
PS: I have rewritten this post, originally starting as "An explicit solution ...",
in order to illustrate the perhaps counterintuitive result of Myerson's Monthly paper.
A: $1_\mathbb{Q}$ is a "double limit" of continuous functions in the sense that we define $f_{nm}(x) = \cos(n!\pi x)^{2m}$ which converges pointwise to $1_\mathbb{Q}$ for $n,m \to \infty$.
Define $g_{nm}(x) = -\log(1-f_{nm}(x))$ which seems to do (close to) what you want.
Only problem is that there might not be a clever way to run through $\mathbb{N} \times \mathbb{N}$ to get a single sequence such that it converges pointwise to $0$ for each irrational $x$.
