Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $X$ is a normal topological space. Suppose some metric space for example. If $\{A_n\}_{n=1}^{\infty}$ is a collection of pairwise disjoint closed subsets of $X$, can we find a continuous function on $X$ such that it takes the constant value $n$ on $A_n$?

share|improve this question
    
I think you should modify the question. It is too easy to give counterexamples. What about: does there exist a continuous function $f$ from $X$ to $[0,1]$ which is constant equal to $a_n$ on each $A_n$, with $a_n\neq a_m$ for all $n\neq m$? Brian M. Scott's answer gives a sufficient condition for this to hold. –  1015 Mar 17 '13 at 20:16
add comment

1 Answer

You need the collection $\mathscr{A}=\{A_n:n\in\Bbb Z^+\}$ to be discrete, meaning that each $x\in X$ has an open nbhd $U$ that meets at most one of the sets $A_n$. Then it’s possible to find a discrete collection $\mathscr{U}=\{U_n:n\in\Bbb Z^+\}$ of pairwise disjoint open sets such that $A_n\subseteq U_n$ for each $n\in\Bbb Z^+$.

Proof. For $n\in\Bbb Z^+$ let $B_n=\bigcup_{k>n}A_k$; since $\mathscr{A}$ is discrete, each $B_n$ is closed. $X$ is normal, so there are disjoint open sets $G_1$ and $H_1$ such that $A_1\subseteq G_1$ and $B_1\subseteq H_1$. Similarly, there are disjoint open sets $G_2$ and $H_2$ such that $A_2\subseteq G_2$ and $B_2\subseteq H_2$, and we may clearly assume that $G_2\cup H_2\subseteq H_1$. Continue in the same fashion: at stage $n>1$ choose disjoint open sets $G_n$ and $H_n$ such that $A_n\subseteq G_n\subseteq H_{n-1}$ and $B_n\subseteq H_n\subseteq H_{n-1}$. Then $\{G_n:n\in\Bbb Z^+\}$ is a pairwise disjoint family of open sets such that $A_n\subseteq G_n$ for each $n\in\Bbb Z^+$.

Now let $C=X\setminus\bigcup_{n\in\Bbb Z^+}G_n$ and $A=\bigcup_{n\in\Bbb Z^+}A_n$; $A$ and $C$ are disjoint closed sets in $X$. If $C=\varnothing$, $\{G_n:n\in\Bbb Z^+\}$ is an open partition of $X$ and hence a discrete collection, so we just set $U_n=G_n$. Assume, then, that $C\ne\varnothing$, and let $V$ and $W$ be disjoint open sets such that $A\subseteq V$ and $C\subseteq W$. For each $n\in\Bbb Z^+$ let $U_n=G_n\cap V$; I claim that $\mathscr{U}=\{U_n:n\in\Bbb Z^+\}$ is discrete.

To see this, suppose that $x\in X$. If $x\in C$, then $W$ is an open nbhd of $x$ that meets none of the sets in $\mathscr{U}$, so suppose that $x\notin C$. Then $x\in G_n$ for some $n\in\Bbb Z^+$, and $U_n$ is the only member of $\mathscr{U}$ that $G_n$ meets. Thus, every point of $X$ has an open nbhd meeting at most one member of $\mathscr{U}$, which is therefore a discrete family. $\dashv$

For each $n\in\Bbb Z^+$ let $f_n:X\to[0,1]$ be a continuous function such that $f_n(x)=1$ if $x\in A_n$, and $f_n(x)=0$ if $x\in X\setminus U_n$. Now let $f=\sum_{n\in\Bbb Z^+}nf_n$, and verify that $f$ has the desired properties; continuity follows from the fact that the family $\mathscr{U}$ is discrete.

If $\mathscr{A}$ is not discrete, there is an $p\in X$ such that every open nbhd of $p$ meets infinitely many members of $\mathscr{A}$, and it’s clear that if a function $f$ satisfies $f(x)=n$ for $x\in A_n$ for all $n\in\Bbb Z^+$, then $f$ cannot be continuous at $p$.

share|improve this answer
    
Ok, let me try again. What if we simply ask, more generally, that there exists $f=\sum_{n\geq}f(n)1_{A_n}$ injective with values in $[0,1]$ that extends continuously to $X$? I think that would be a proper generalization of Urysohn. I couldn't find a counterexample. Any suggestion? –  1015 Mar 17 '13 at 20:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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