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In an arbitrary metric space, does it follow that if $x \ne y$ and $\epsilon \neq \delta$ then $B_\epsilon(x) \ne B_\delta(y)$?

I would say this is not true. We can find a counterexample in the $(X, d_0)$, the discrete metric space. If we take $B_\epsilon(x)$ and $B_\delta(y)$ such that $x \ne y$ and $\epsilon \neq \delta$ with both $\epsilon$ and $\delta$ greater than $1$, then these open balls will be equal as they will both contain all points of $X$. Have I got that right?

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That's correct. –  Austin Mohr Nov 25 '12 at 2:23

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up vote 1 down vote accepted

Yes.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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