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True or false: The closed balls of a metric space are precisely those subsets such that every proper superset has strictly greater diameter.

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you mean smaller? –  mathemagician Jan 19 '13 at 15:28
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To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. –  Zev Chonoles Jan 19 '13 at 15:28
    
Is this a question, a challenge, or what? –  Matemáticos Chibchas Jan 19 '13 at 15:46
    
mathemagician: No I think greater is right. Zev: It's not homework, it's a conjecture. I'm really struggling - to get started, I've been trying to show that the diameter of every ball in a metric space is 2 times its radius, without success. I only just realized that this is in fact false. Matematicos: It's a genuine question - I wish to see a counterexample, or else some ideas about how to prove it. –  goblin Jan 19 '13 at 15:51
    
I think that any closed convex set will satisfy this. –  Tomás Jan 19 '13 at 15:59

1 Answer 1

up vote 4 down vote accepted

False: take $X=\{(-2,0),(0,0),(2,0),(0,3)\}$ in $\mathbb R^2$ with the induced metric. Then the closed ball centered at $(0,0)$ of radius $2$ is the set $B=X\setminus\{(0,3)\}$, and we have $\mathrm{diam}\ X=\mathrm{diam}\ B=4$.

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