In a Completely regular $T_1$ space, two disjoint sets, one compact, the other closed, can be separated by a continuous function?

Question

Let $X$ be a completely regular $T_1$ space and let $A,B$ be disjoint closed subsets of $X$, where $A$ is compact also. Then is it true that there exist a continuous function $f\colon X \to [0,1]$ such that $f(A)=\{0\}$, $f(B)=\{1\}$?

Answer 1

HINT: Take for every $a$ in $A$ a continuous function $f_a\colon X \to [0,1]$ such that $f_a(a) = 0$ and $f_a\ _{|B} = 1$. The open subsets $U_a \colon = f_a^{-1}([0, \frac{1}{2}))$ form an open cover of the compact subset $A$ so there exist finitely many $a_1$, $\ldots, a_m $ so that $\cup_{i=1}^m U_{a_i}$ cover $A$. Consider the continous function $g\, \colon= \min (f_1, \ldots, f_m)$. It take values in $[0,1]$, is $< \frac{1}{2}$ on $A$ and takes value $1$ on $B$. We are almost done. Compose $g$ with a function $\phi$ from $[0,1]$ to $[0,1]$ that is $0$ on $[0,\frac{1}{2}]$ and $1$ at $1$ ( say $\phi(t) = \max \{ 0, 2t-1 \}$ ) to get $f= \phi \circ g$ that separates $A$ and $B$.