# Application of Urysohn's Lemma to non-disjoint closed sets

Let $X$ be a normal space with the property that every closed set in $X$ is a countable intersection of open sets in $X$. Then show that:

(a) Given $A \subset X$ closed, $\exists$ a continuous map $f : X \rightarrow [0,1]$ such that $f^{-1}(0) = A$.

(b) Given $A,B \subset X$ closed, $\exists$ a continuous map $f : X \rightarrow [0,1]$ such that $f^{-1}(0) = A$ and $f^{-1}(1) = B$.

I know I need to use 'Urysohn's Lemma'. But I'm not able to see how to apply.

Thanks for any help.

• Your title mentions non-disjoint closed sets. Of course the sets $A,B$ in (b) need to be disjoint because $f^{-1}(0)\cap f^{-1}(1)=\emptyset$. – Hagen von Eitzen Apr 17 '15 at 14:00
• Oh yes, I'm sorry. – Richard K. Apr 17 '15 at 14:02

(a) Let $U_n$, $n\in \mathbb N$ be open sets with $A=\bigcap_{n\in\mathbb N} U_n$. Then $X-U_n$ is closed and disjoint from $A$, hence there exists continuous $f_n\colon X\to[0,1]$ with $A\subseteq f_n^{-1}(0)\subseteq U_n$. Now let $$f(x)=\sup\{\,\tfrac1nf_n(x)\mid n\in\mathbb N\,\}.$$ Then $f$ is a function $X\to [0,1]$ with $A\subseteq f^{-1}(0)\subseteq f_n^{-1}(0)\subseteq U_n$, hence $A=f^{-1}(0)$ by taking intersection over all $n$. Remains to show that $f$ is continuous, that is: $f^{-1}((a,1])$ is open for $0\le a<1$, and that $f^{-1}([0,a))$ is open for $0<a\le 1$. The first follows from $f^{-1}((0,1])=X-A$ for $a=0$ and from $$f^{-1}((a,1])=\bigcap_{n\in\mathbb N} f_n^{-1}((na,1])=\bigcap_{1\le n<1/a} f_n^{-1}((na,1])$$ for $a>0$. The second follows from $$f^{-1}([0,a))=\bigcap_{n\in\mathbb N} f_n^{-1}([0,na))=\bigcap_{1\le n\le 1/a+1} f_n^{-1}([0,na)).$$
(b) Apply (a) to $A$ and $B$, giving functions $f_A, f_B$. Now let $$f(x)=\frac{f_A(x)}{f_A(x)+f_B(x)}.$$ For this to be defined, we need that $f_A(x)+f_B(x)>0$ for all $x$, which is equivalent to $A\cap B=\emptyset$, a necessary (and obviously so) condition missing from the problem statement!