1
$\begingroup$

Let $E$ be a compact metric space. Suppose that closure of every open ball $B(a,r)$ is the closed ball $B'(a,r)$. Must every open ball in $E$ be connected? I think it most probably is. But I don't know how to go about proving this.

New contributor
cherry is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

put on hold as off-topic by user44191, Sean Lawton, Yemon Choi, Davide Giraudo, Wojowu Mar 15 at 17:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – user44191, Sean Lawton, Yemon Choi, Davide Giraudo, Wojowu
If this question can be reworded to fit the rules in the help center, please edit the question.

3
$\begingroup$

Yes, every open ball is connected.

Suppose the open ball $B(a,r)$ is disconnected: $B(a,r) = U \cup V$ where $U$ and $V$ are nonempty, open and disjoint, and $a \in U$. Since $\overline{V}$ is compact, there is a point $v \in \overline{V}$ whose distance $s = d(a,v)$ to $a$ is minimal. Since $V \subset B(a,r)$, $s < r$ and $v \in B(a,r)$. Note that $U \cap \overline{V} = \overline{U} \cap V = \emptyset$, so $v \in V$ and $v \notin \overline{U}$. Thus we have $v \in B'(a,s)$, but $B(a,s) \subseteq U$ so $v \notin \overline{B(a,s)}$, contradicting the assumption $\overline{B(a,s)} = B'(a,s)$.

$\endgroup$
  • $\begingroup$ Can you explain how v does not belongs to closure of U. I don't think anything is stopping from this. Of course then s would be zero. $\endgroup$ – cherry Mar 16 at 6:37
  • $\begingroup$ @cherry $v$ is in $V$, which is open. Thus there is a neighbourhood of $v$ which is disjoint from $U$, so $v \notin \overline{U}$. $\endgroup$ – Robert Israel 2 days ago
1
$\begingroup$

Yes, it is true.

Assume that an open ball $B(a,R)$ is not connected. Let $S\not\ni a$ be a connected component of $B(a,R)$. Since the space is compact there is a point $s\in S$ that minimize the distance $|a-s|$. Note that $s$ does not lie in the closure of $B(a,r)$ for $r=|a-s|$ --- a contradiction.

$\endgroup$
  • $\begingroup$ Why is $S$ compact (from which you know distance from $S$ to $a$ has a minimum)? A connected component of an open subset is closed in the open subset but need not be closed (hence compact) in the whole space, in general. $\endgroup$ – KConrad Mar 15 at 10:01
  • 1
    $\begingroup$ @KConrad: I don't think this is being claimed: I interpret "the space" as the whole space, not $S$, though then the claim should be "... there is a point $s\in\overline{S}$ ..." Robert just posted the same argument with more details. $\endgroup$ – Christian Remling Mar 15 at 14:57
  • 1
    $\begingroup$ Oh, so "$s \in S$" should be "$s \in \overline{S}$". Then my objection no longer applies. $\endgroup$ – KConrad Mar 15 at 15:07
  • $\begingroup$ @KConrad, Note that $|x-a|=R$ for any $x\in \bar S\backslash S$, therefore $s\in S$. $\endgroup$ – Anton Petrunin Mar 16 at 0:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.