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If in a metric space we have $B(x,r) = B(y, s)$, is it necessary that $x = y$ and $r = s$?

I think that the center of the balls i.e. $x$ and $y$ must be same but the radius $r$ and $s$ may not be same.....and then also the balls may be same.

For example in discrete metric space $B(x,1/2)$ and $B(x, 3/4)$ are same!!!

Am I correct??

User8976
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2 Answers2

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In $\mathbf{Z}$ with the classic (discrete) topology, any ball of radius $<1$ is equal to its center, so your right. But in the $p$-adic field $\mathbf{Q}_p$ for instance, any point of any ball (closed or not) is its center, and balls are either disjoint, or one in the other. But it $\mathbf{R}$ with the usual distance, it is necessary. So... it depends ;-)

Advice : You can think of a metric space as something like $\mathbf{R}$ when you begin with topology of metric spaces, everyone did this, but be aware that there are things like $\mathbf{Q}_p$ etc, but you spotted it alone ! ;-)

Olórin
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  • Let $p$ be a positive prime number. For $x\in\mathbf{Q}$, define $|x|_p = 0$ if $x=0$, and as $|x|_p = p^{-d}$ if $x\not=0$ when $x = p^d \frac{r}{s}$ with $d,r,s\in\mathbf{Z}$, $r\wedge s = 1$ and $p\not| r$, $p\not|s$. ($d$ is the $p$-adic valuation of $x$.) You can see that $|\cdot|_p$ defines a distance on $\mathbf{Q}$ such that $|x-y|_p \leq \max(|x|_p, |y|_p)$ for each $x,y$. This special property is coined as "ultrametric". Then $\mathbf{Q}_p$ is the completion of $\mathbf{Q}_p$ for the distance $|\cdot|_p$. – Olórin Jan 21 '15 at 02:18
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As a counterexample, how about the simple metric space consisting of only two points $x$ and $y$, and $d(x,y)=1$. Then $B(x,2)=B(y,3)$, because both balls are equal to the entire space $\{x,y\}$.

Mankind
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