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Assuming that $d(x,x') \ge r+r'$, and proving that they are disjoint is easy. It's the other side that I'm having difficulty with. This seems like a really easy problem, but i'm having difficulty proving it with basically just using the triangle inequality. Any help appreciated!

Michael Hardy
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Tejus
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    You need $\ge$, not $\le$. ${}\qquad{}$ – Michael Hardy Apr 03 '14 at 16:55
  • It's plain false. – user2345215 Apr 03 '14 at 16:59
  • Somebody down-voted this. An explanation for that would help. @user2345215 : Yes, it's "plain false", but "$\le$" rather than "$\ge$" was probably a typo. It's true in Euclidean spaces (modulo the typo), which I suspect were clear in the context in which the problem was stated. – Michael Hardy Apr 03 '14 at 17:13
  • @MichaelHardy It's plain false even after considering the possible typo. The question says nothing about Euclidean spaces and the tag is metric spaces. Hence -1. – user2345215 Apr 03 '14 at 17:17
  • That someone has identified the fact that it's hard to prove without additional assumptions is worth a "+1". ${}\qquad{}$ – Michael Hardy Apr 03 '14 at 17:24
  • Thanks so much! yes, it was a typo. Just corrected. Apologies for the sloppiness. Thank you so much for the answer. You're right, the question was originally framed within Euclidean spaces (where it can easily be seen to be true). However, I was thinking that it should also work in a metric space setting, which clearly isn't the case. – Tejus Apr 03 '14 at 17:42

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It is false in metric spaces in general. For example, suppose only two points exist in the space and the distance between them is $1$. Then the open balls of radius $3/4$ about the two points are disjoint even though $3/4+3/4>1$.

So you need some additional assumptions about the space involved beyond what things like the triangle inequality can give you.

If it's a Euclidean space, you can look at the set $\{wx+(1-w)x' : 0\le w\le1\}$, which is just the segment with endpoints $x$ and $x'$. If the two open balls are disjoint, you should be able to find a point $y$ on that segment whose distance from $x$ is more than $r$ and whose distance from $x'$ is more than $r'$, and you should be able to show that the distance from $x$ to $y$ plus the distance from $y$ to $x'$ is the distance from $x$ to $x'$.

Michael Hardy
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