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Let $M$ be a non-empty metric space with the following property: if $B(x,r) = B(y,s)$, then $x=y$ and $r=s$, where $B(x,r)$ is the closed ball with center $x$ and radius $r$. Must $M$ be infinite? In fact, must $M$ be uncountable?

user107952
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    If there is even one $x$ such that the balls $B(x,r)$ for $r=1,2,3,\dots$ are all distinct, then $M$ must be infinite. On the other hand, the space of rational numbers with the usual metric has your property but is countable. – bof Dec 11 '20 at 00:43
  • @bof That should be an answer. – Noah Schweber Dec 11 '20 at 05:49

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Uncountable, no. The space of rational numbers with the usual metric has your property but is countable.

Infinite, yes. If $B(x,1),B(x,2),B(x,3),\dots$ are all different then $M$ must be infinite, since for each $n\in\mathbb N$ there is a point $y_n$ such that $d(x,y_n)\in(n,n+1]$.

bof
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  • What if the radius is irrational? Then, does the space of rational numbers still have the property? – user107952 Dec 11 '20 at 21:11
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    Yes. For instance, if $r,s\in\mathbb R$, $0\lt r\lt s$, then $\mathbb Q\cap(-r,r)$ is a proper subset of $\mathbb Q\cap(-s,s)$, because there are rational numbers between $r$ and $s$; it does not matter if $r$ and $s$ are rational or irrational. – bof Dec 11 '20 at 22:11