October 19th, 2013, 07:41 PM  #1 
Newbie Joined: Sep 2013 Posts: 8 Thanks: 0  Disjoint open balls
Trying to prove =\emptyset \leftrightarrow r+r'\leq d(x,x'" /> I can go the direction but am having a very hard time proving would really appreciate some help, thank you 
October 20th, 2013, 12:26 AM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Disjoint open balls
It's not generally true, so you're going to need some other property of the metric space. E.g. it's not necessarily true if the metric space is disconnected. E.g. [0,1] U [2,3], consider B(1,3/4) and B(2, 3/4). It's only going to be true if you have the existence of points all the way between x and x'. For IR^n, you could consider the point x + r(x'x)/d(x, x'). 
October 20th, 2013, 07:07 AM  #3 
Newbie Joined: Sep 2013 Posts: 8 Thanks: 0  Re: Disjoint open balls
Thank you for responding.. sorry for not being specific, we're in R^n here. In what way could I consider that point when trying to prove the statement? Im not having any trouble seeing why this statement is true intuitively, formulating a formal proof is what I am finding difficult. In the left direction It's easy to assume the right side and directly show that an element of one ball is not an element of the other ball, but going the right direction I don't know where to begin. Are you suggesting I use the point you mentioned to do this? 
October 21st, 2013, 12:12 AM  #4 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Disjoint open balls
First, consider closed balls, where you have strict inequality: " /> , \ then \ y = x + \frac{r(x'x)}{d(x,x'} \in B[x,r]\cap B[x',r']" /> And, in fact, if you have equality then y is the only point in the intersection. Think vectors! For open balls, you need to change r to ra,where 0 < a < (r+r')  d(x,x'). 

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balls, disjoint, open 
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