If $X$ is a Hausdorff topological space, we say that $X$ is perfectly normal if, for every closed set $A$, there exists a continuous function $f_A:X\rightarrow[0,1]$ such that $f_A^{-1}(0)=A$.
Prove that perfectly normal spaces are normal.
My effort:
Suppose $A$ and $B$ are two disjoint closed set. Define $g=\frac{f_A}{f_A+f_B}$. $g(x)=0$ if and only if $x\in A$, and $g(x)=1$ if and only if $x\in B$.
As long as I can prove $g:X\rightarrow [0,1]$ is continuous, I am done. But I do not know how to show it is continuous.
Best Answer
Another way can be to see your map $g$ as the following composition
$g=G\circ (f_A,f_B)$ where
$(f_A,f_B): X\to [0,1]\times [0,1]$
and
$G: E\subseteq [0,1]\times [0,1]\to [0,1]$
where $E=\{(x,y): x\neq 0 \textit{ or } y\neq 0\}$
And the map is $G(x,y):= \frac{x}{x+y}$
that is continuos on its domain. This map is continuos for the same reasoning of Kavi Rama Murthy.