Let if $B_r(x)$=$B_s(y)$ for some $x$,$y$ in metric space $M$ and $r$,$s$ $\in$ $R$. Is true $x=y$? Is true $r=s$?
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No. For example:
$$X = \{ 1,2\}\\ d(x,y) = 0 \text{ if $x=y$, 1 otherwise}\\ x = 1, y= 2, r=16, s=26$$
mookid
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1This makes me wonder if it is true in general for non-discrete spaces. Can you think of a counter example? – Cameron Williams Mar 07 '14 at 22:40
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2It is true for normed vectorial spaces. – mookid Mar 07 '14 at 22:41
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Yeah that's definitely the case. Good point. – Cameron Williams Mar 07 '14 at 22:42
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What? $B_1(x)={x}$, and $B_2(y)={x,y}$ here, so it's not a counterexample (surely we're working with open balls?) It works however if you take $r=42$, $s=9001$. Also, you have a typo in your def: you want $x \neq y$, not $x = y$. – Joshua Pepper Mar 07 '14 at 22:47
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I considered closed balls. simply take 2 and 3 in the case of open balls :) and thank you for the more serious mistake. – mookid Mar 07 '14 at 22:50
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Another counterexample. Take any bounded metric space of diamter $r$. Any ball of radius greater than $\frac{r}{2}$ will cover the entire space and so all such balls are equal no matter their center.
Dan Rust
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@mookid I agree, though I think my answer possibly has more explanatory power. It's also then not hard to imagine examples where you don't need to cover the entire space or even use bounded spaces, such as $X=[0,\infty)$ and the balls $B_1(0)$ and $B_{3/4}(\frac{1}{4})$. – Dan Rust Mar 07 '14 at 22:55
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Building on your idea, a similar thing can happen if the space is disconnected, but it may be that only a limited range of radii and centers will yield equal balls. Consider $X=[0,1]\cup[2,3]$. – MPW Mar 08 '14 at 02:58