Does there exist a connected metric space with more than one point such that it has an open ball which is not connected ? Moreover does there exist a connected metric space with more than one point such that every open ball is not connected ? Please help . Thanks in advance .

  • 1
    $\begingroup$ ...why did you ask the exact same question twice? Anyway, any thoughts on that question? $\endgroup$ Sep 4, 2015 at 9:15
  • $\begingroup$ @NajibIdrissi : I have edited $\endgroup$
    – user228168
    Sep 4, 2015 at 9:23

3 Answers 3


How about we take the following subset of $\mathbb{R}^2$: $$ X = \mathbb{R} \times \{0\} \cup \mathbb{R} \times \{1\} \cup \{0\} \times [0,1] $$ That is two parallel lines joined by a line perpendicular to them. Now take any point on one of those lines which is far away from this connecting line (for instance $(3,0)$ would do) and take the open ball with radius $2$ of it. This will be the disjoint union of nonempty subsets of $\mathbb{R} \times \{0\}$ and $\mathbb{R} \times \{1\}$ and hence disconnected.


I’ll show that Crostul’s space is indeed an example of a connected metric space in which no open ball is connected.

Let $f:\Bbb Z\to\Bbb Q$ be a bijection, and let

$$X=\big((\Bbb R\setminus\Bbb Q\big)\cup\{\langle f(n),n\rangle:n\in\Bbb Z\}$$

as a subspace of $\Bbb R^2$.

Let $H$ be a non-empty clopen subset of $X$. If $x\in\Bbb R\setminus\Bbb Q$, then $H_x=\{y\in\Bbb R:\langle x,y\rangle\in H\}$ is a clopen subset of $\Bbb R$, so $H_x=\varnothing$ or $H_x=\Bbb R$. Let $A=\{x\in\Bbb R\setminus\Bbb Q:H_x=\Bbb R\}$; clearly $H\supseteq A\times\Bbb R$, and $H\cap\big((\Bbb R\setminus A)\times\Bbb R\big)=\varnothing$.

Suppose that $\langle f(n),n\rangle\in H$ for some $n\in\Bbb Z$; then there is an open nbhd $V$ of $f(n)$ in $\Bbb R$ such that $(V\setminus\Bbb Q)\times\{n\}\subseteq H$, so $f(n)\in\operatorname{cl}_{\Bbb R}A$. Conversely, if some $f(n)\in\operatorname{cl}_{\Bbb R}A$, then $\langle f(n),n\rangle\in\operatorname{cl}_XH=H$. Thus, $H=X\cap\big((\operatorname{cl}_{\Bbb R}A)\times\Bbb R\big)$. But

$$X\setminus H=X\cap\big((\Bbb R\setminus\operatorname{cl}_{\Bbb R}A)\times\Bbb R\big)$$

is also clopen, so essentially the same argument shows that $\Bbb R\setminus\operatorname{cl}_{\Bbb R}A$ must be closed in $\Bbb R$, which means that it must be empty, i.e., that $H=X$ and hence that $X$ is connected.

Now let $p=\langle x,y\rangle\in X$ and $r>0$ be arbitrary, and let $B=B(p,r)$, the open ball of radius $r$ centred at $p$. Choose $m\in\Bbb Z^+$ so that $m>\max\{|y-r|,|y+r|\}$. $\Bbb Q\cap(x-r,x+r)$ is infinite, so there is an $n\in\Bbb Z$ such that $|n|\ge m$, and $f(n)\in(x-r,x+r)$. Then $\langle f(n),n\rangle\notin B$, so $\{\langle u,v\rangle\in B:u<f(n)\}$ is a non-empty, clopen, proper subset of $B$, which is therefore not connected.


I would try with this:

Firstly, fix a bijection $f: \Bbb{Z} \longrightarrow \Bbb{Q}$. Then, define the following subspace of $\Bbb{R}^2$ $$X=((\Bbb{R} \setminus \Bbb{Q}) \times \Bbb{R} ) \cup \{ (f(n), n) : n \in \Bbb{Z} \}$$

I think that this is an example of a connected metric space where all balls are disconnected. However this is just intuition, I have no proofs about this.

  • $\begingroup$ I added a proof that your example works. $\endgroup$ Sep 4, 2015 at 22:42
  • $\begingroup$ @BrianM.Scott Thanks! $\endgroup$
    – Crostul
    Sep 5, 2015 at 17:20
  • $\begingroup$ You’re welcome! Good intuition on your part. $\endgroup$ Sep 5, 2015 at 17:26

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