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Does there exist a connected metric space, with more than one point and without any isolated point, in which at least one open ball is countable?

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Assume a ball $B(x_0,r)\subseteq X$ is countable, $x_0\in X$, $r>0$. As $X$ is not a one-point space, there exists $x_1\ne x_0$ and we may assume $r\le d(x_0,x_1)$. Then by pigeon-hole there exists $0<\rho <r$ such that $d(x_0,x)\ne \rho$ for all $x\in B(x_0,r)$. By definition of $B(x_0,r)$, in fact $d(x_0,x)\ne\rho$ for all $x\in X$. Then the sets $\{\,x\in X:d(x_0,x)<\rho\,\}$ and $\{\,x\in X:d(x_0,x)>\rho\,\}$ are open, disjoint, cover $X$ and are not empty (as witnessed by $x_0$ and $x_1$, respectively. Hence $X$ is not connected.

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  • $\begingroup$ @ Hagen von Eitzen : Why does there exist $0<\rho<r$ such that $d(x_0 , x) \ne \rho$ ? and how $d(x_0 , x) \ne \rho , \forall x \in X$ ? and are we anywhere using that $X$ has no isolated point ? $\endgroup$
    – user228168
    Jun 16 '15 at 13:23
  • $\begingroup$ @SuanDev no isolated points for a connected space just means it must have more than one point, so yes it is being used. $\endgroup$ Jun 16 '15 at 13:26
  • $\begingroup$ @Hagen von Eitzen : Ok , I have understood :) $\endgroup$
    – user228168
    Jun 16 '15 at 13:42

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