A topology product problem with sequence of continuous and unbounded functions

Question

Consider $X=\{f:\mathbb{Q}\rightarrow \mathbb{R}\}$ (set of any functions), with the product topology. Is true that for any $f\in X$, there are $(f_n)_{n\in \mathbb{N}}$ sequence of continuous and unbounded functions such that $f_{n}\rightarrow f$?

I tried to solve that problem using the usual notion of point convergence in $\mathbb{Q}$ and I discovered that is possible for a function with a discrete set of discontinuous points. In this case I construct a sequence with a asymptotic behavior near these points, but my proof was very complicated and I can't find a proof in the general case, which requires the product topology.

Answer 1

Let $(q_n)_{n\in\Bbb N}$ be an enumeration of $\Bbb Q$. For each $n\in\Bbb N$, let $\{r_1,r_2,\ldots,r_n\}$ the set $\{q_1,q_2,\ldots,q_n\}$ , with $r_1<r_2<\cdots<r_n$. Let $f_n\colon\Bbb Q\longrightarrow\Bbb R$ be the function such that:

• if $x\leqslant r_1$, then $f_n(x)=x-r_1+f(r_1)$;
• if $x\geqslant r_n$, then $f_n(x)=x-r_n+f(r_n)$;
• otherwise, take $k\in\{1,2,\ldots,n\}$ such that $r_{k-1}\leqslant x\leqslant r_k$ and define$$f_n(x)=\frac{f(r_k)-f(r_{k-1})}{r_k-r_{k-1}}x+\frac{r_kf(r_{k-1})-r_{k-1}f(r_k)}{r_k-r_{k-1}}$$and note that $f_n(r_{k-1})=f(r_{k-1})$ and that $f_n(r_k)=f(r_k)$.

So, $f_n$ is continuous and unbounded. Furthermore, $f_n(q_k)=f(q_k)$ for each $k\leqslant n$. So, $(f_n)_{n\in\Bbb N}$ converges pointwise to $f$.