Can a sequence of functions have infinity as limit exactly at rationals?

Since there is a homeomorphism of $\Bbb R $ taking $\Bbb Q $ to the dyadic rationals, that is, rational numbers where the denominator is a power of 2, (see here for an example of such a function), we can restrict out attention to finding a sequence of functions converging to infinity exactly at the dyadics.

Now for $ p, d \in \Bbb R $ we let $ h_{p, d} $ denote a hat function, i. e. a continuous function with $ h_{p, d}(p) =1$ and $ h_{p, d}(x) =0$ for $ x \notin [p-d, p+d]$.

Now define $ f_n = \sum_{k \in \Bbb Z} n h_{k/2^n,2^{-n-2} }$

It is obvious that for dyadic x, $ f_n (x) \to \infty $ as $ n \to \infty $.

If $x$ is not dyadic, there are infinitely many $ n $ with $\{2^n x\} \in [1/4,1/2]$ (here the brackets denote the fractional part of the real number). To see this, consider the binary expansion of $ x $ and note that, since this expansion is not allowed to end in an infinite chain of zeros or in an infinite chain of ones, there are infinitely many integers $ n $ such that the $ n+1$-th digit after the dot is a $0$ and the $ n+2$-th is a $1$.

But with such a choice of $ n $, we get $ f_n (x)=0$ which shows that $ f_n (x )\not \to \infty $ as $ n \to \infty$.