$X$ is a normal Hausdorff space and $A,B,C$ three pairwise disjoint closed sets. I want to prove the existence of a continuous real-valued function $f$ that takes the values $a,b,c$ on $A,B,C$ respectively. Is this a direct application of Urysohn's lemma?
What I tried :
$A$ and $B\cup C$ are disjoint closed sets. So $\exists$ disjoint open sets $U,V$ such that $A\subset U, B\cup C\subset V$. Again $\exists$ disjoint open sets $V_1,V_2$ such that $B\subset V_1, C\subset V_2$. Taking their intersections with $V$ I obtain pairwise disjoint open sets $U_1,U_2,U_3$ with $A\subset U_1, B\subset U_2, C\subset U_3$. And that's all I could think of unfortunately.
Any help would be greatly appreciated.
Best Answer
By Uyrsohn's Lemma applied to $A$,$B \cup C$ resp. $B$,$C$ there exist continuous $f:X \to \Bbb R$, $g:X \to \Bbb R$ with:
$$f[A] = \{0\}, f[B \cup C] = \{1\}, g[B] = \{0\}, g[C] = \{1\}$$
Define $h(x) = f(x)(g(x) + 1)$, which is continuous and satisfies:
$$h[A] = \{0\}, h[B] = \{1\}, h[C] = \{2\}$$
This generalizes to any finite number of pairwise disjoint sets.
$$a + f(x)\left((c - b)g(x) + b - a\right)$$ gives $a,b,c$ on $A$, $B$ and $C$ respectively.