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Is it true that in an infinite metric space, any open ball of radius $2$ is an infinite set?

for example $\mathbb{R}^2$ with discrete metric we have $d(x,y)=1\forall x\ne y$ so in this case also we have whole $\mathbb{R}^2$ within a ball of radius $2$ right? is my concept okay?

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    $\begingroup$ Obviously not. Take any infinite set with the metric $d(x,y) = 2\delta_{xy}$. There, every ball of radius $2$ is not just finite, it's a singleton. $\endgroup$ – kahen Jun 9 '13 at 16:20
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What do you mean by "infinite metric space"? Take $\Bbb Z^2$ with the usual Euclidean metric, and you can see the $2$-ball centered at $0$ contains finitely many elements.

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No. If you start with a discrete space instead of the discrete metric, you can find plenty of counter examples. Take $\mathbb{Z}$ with its subspace topology, and any finite-radius open ball, for instance.

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Consider the metric on $\mathbb{R}$ defined by

$$d(x,y) = \begin{cases} 5 \text{ if } x \neq y & \\ 0 \text{ if } x = y \end{cases}.$$

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