I'm having some difficulty with the following problem in general topology:

Prove that if the closure of each open ball in compact metric space is the closed ball with the same center and radius, then any ball in this space is connected.

I've tried many things - looking at the components of the open ball, the closed ball, assuming there is a non constant function from the open\closed ball to $\{0,1\}$ & more - But I always ended up with more mess then I can handle.

Any ideas?


Assume $X$ is disconnected, i.e. the disjoint union of two closed sets $A$ and $B$. Note that $B$ is compact. For any point $a\in A$, there is thus a point $b\in B$ such that $d(a,b)=r$ is the largest radius of a ball $B_r(a)$ contained in $A$. Can you finish from here?

  • $\begingroup$ Yes, I think I can. $A$ is compact as well and $\cup _{a \in A} B_{r_a}(a)$ is an open cover of $A$, thus has a finite subcover $A = \cup_{i=1}^n B_{r_i}(a_i)$. $A$ is closed so $A = \bar A = \cup_{i=1}^n \bar B_{r_i}(a_i)$. But $\bar B_{r_i}(a_i)$ contains an element from $B$, since as you said, there is a $b \in B$ such that $d(a_i, b) = r_i$. Contradiction. So this shows that $X$ is connected. $\endgroup$ – amirbd89 Jul 16 '15 at 21:53
  • $\begingroup$ Every closed ball is also connected since as a subspace of X it inherits the 2 properties presented in the question: It is compact and every open ball's closure is the closed ball. Since every closed ball is connected and every open ball is the union of closed balls that doesn't have empty intersection pairwise, then every ball is connected. Thank you. $\endgroup$ – amirbd89 Jul 16 '15 at 21:56
  • $\begingroup$ I just realized that the first part can be done much quicker : I don't need to cover $A$ at all, It's enough to find one $a$ and $r_a$. Then since $B_{r_a}(a) \subseteq A$ then $\bar B_{r_a}(a) \subseteq \bar A = A$, and we've shown that $\bar B_{r_a}(a)$ contains an element from $B$ and that's of course a contradiction. $\endgroup$ – amirbd89 Jul 17 '15 at 0:16
  • $\begingroup$ @amirbd89: Right $\overline B_r(a)$ is contained in $A$. But if we assume that this is the closed ball of radius $r$, then since $d(a,b)=r$, it contains $b$, a contradiction. $\endgroup$ – Stefan Hamcke Jul 17 '15 at 1:54

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