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October 19th, 2013, 08:41 PM   #1
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Disjoint open balls

Trying to prove
=\emptyset \leftrightarrow r+r'\leq d(x,x&#39" />
I can go the direction but am having a very hard time proving
would really appreciate some help, thank you
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October 20th, 2013, 01:26 AM   #2
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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').
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October 20th, 2013, 08:07 AM   #3
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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?
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October 21st, 2013, 01:12 AM   #4
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Re: Disjoint open balls

First, consider closed balls, where you have strict inequality:

" />

, \ then \ y = x + \frac{r(x'-x)}{d(x,x&#39} \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 r-a,where 0 < a < (r+r') - d(x,x').
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