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.

I'm trying to show:

Let $(X,d)$ be a metric space and let $A, B$ be nonempty subsets, which are also closed and disjoint. Let $\rho_A:X\to \mathbb{R}$ be such that $\rho_A=d(x,A)$ and $\rho_B:X\to \mathbb{R}$ be such that $\rho_B=d(x,B)$, with $x\in X$ (distance from one point to a set).

Prove that the function $\frac{\rho_A}{\rho_A+\rho_B}:X\to \mathbb{R}$ is continuous in $X$.

I have only the definition of continuous function (with balls) and some results. I can not use yet sequences.

A corollary says that composition of continuous functions is a continuous function on metric spaces. Now, the sum of continuous functions is continuous (in the metric space $\mathbb{R}$) but the division of continuous functions is not necessarily a continuous function in $\mathbb{R}$.

Any help? Thank you very much.

share|improve this question
1  
First you should show that the function is well-defined, that is, that you are not dividing by zero. –  Rasmus Jun 4 '12 at 14:47
1  
The quotiont of functions is continuous as long as the denominator stays away from zero. So you need to focus on the question whether $\rho_A + \rho_B$ can become zero or not, and if yes, how the function then behaves. –  user20266 Jun 4 '12 at 14:48
    
@Rasmus, just because $A$ and $B$ are not disjoint, I think this is enough. Otherwise fine right? –  Hiperion Jun 4 '12 at 14:51
    
@Hiperion: Disjointess is not enough (but it is required), you have to use the property that $A$ and $B$ are closed. Hint: $\bar{A}=\{x\in X:d(x,A)=0\}$, where $\bar{A}$ is the closure of $A$ in $X$. –  Thomas E. Jun 4 '12 at 14:55
1  
@JasonDeVito: You probably mean that the sum equals $0$? –  Thomas E. Jun 4 '12 at 15:01

1 Answer 1

up vote 2 down vote accepted

You first need to show that the function $\rho_A$ (and $\rho_B$ ) is continuous.

If $A$ is any set, then let $\rho_A(x) = \inf_{a\in A} d(x,a)$. Now suppose $d(x,a) \leq \rho_A(x)+\epsilon$, then for any $y$ we have $d(y,a) \leq d(x,y)+d(x,a) \leq d(x,y)+\rho_A(x)+\epsilon$, from which is follows that $\rho_A(y)-\rho_A(x) \leq d(x,y)+\epsilon$. Reversing the roles of $x$ and $y$ shows that $|\rho_A(x)-\rho_A(y)| \leq d(x,y)+\epsilon$. Since this is true for arbitrary $\epsilon>0$, it follows that the $\rho_A$ is Lipschitz continuous of rank $1$. This is true for any set.

Now suppose $A$ is closed, and $x\notin A$. Then $\exists \epsilon>0$ such that if $d(x,y)< \epsilon$, then $y\notin A$. It follows that $d(x,a) \geq \epsilon$ whenever $a \in A$. Hence $\rho_A(x) \geq \epsilon > 0$.

Since $A$ and $B$ are disjoint, it follows that for any $x$, either $x\notin A$ or $x\notin B$, hence $\rho_A(x)+\rho_B(x) > 0$. Then the function $\frac{1}{\rho_A + \rho_B}$ is continuous. The desired result follows from this.

share|improve this answer
    
Thank you very much. –  Hiperion Jun 5 '12 at 17:38

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.