Let $f:\mathbb{R}\to(\mathbb{R},\mathcal{T})$ be a map defined by $f(x)=x$ if $x\in\mathbb{Q}$, $f(x)=0$ if $x\in\mathbb{R}\setminus\mathbb{Q}$, where $\mathcal{T}$ is the usual topology on $\mathbb{R}$.
Now, suppose that the domain $\mathbb{R}$ is equipped the following topology: $$\mathcal{T}_{w}=\left\{f^{-1}(U)\,:\,U\in\mathcal{T}\right\}.$$ Then, clearly, $f$ is a continuous map.
Is it possible to show the normality of the space $(\mathbb{R},\mathcal{T}_{w})$ using the fact that the continuity of $f$ and the normality of $(\mathbb{R},\mathcal{T})$?
I tried to prove it by definition, but there was no progress.
Can anyone help me? or give some advice? Thank you!
The problem can be re-written as follows: let $F_1,F_2$ be two closed (for the usual topology) subsets of $\mathbb{R}$, such that $F_1 \cap F_2$ has no rational point. Show that there are open subsets $U_i$ such that $U_i \supset F_i \cap \mathbb{Q}$ and $U_1 \cap U_2$ has no rational point, ie $U_1$ and $U_2$ are disjoint.
You can just define, if $i$ is $1$ or $2$, $$U_i=\bigcup_{x \in F_i \cap \mathbb{Q}}{\left(x-\frac{d(x,F_{3-i})}{5},x+\frac{d(x,F_{3-i})}{5}\right)}.$$