0
$\begingroup$

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.

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

(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!

$\endgroup$
  • $\begingroup$ Interesting last point that you made. Really basic and elegant proof. Thanks a lot mate. $\endgroup$ – Richard K. Apr 17 '15 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.