# 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?

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?

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.
• @ 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 ?