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.
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2 Answers
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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 $\overline{B(a,s)} = B'(a,s)$.
answered Mar 15, 2019 at 14:33
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$\begingroup$ 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. $\endgroup$– cherryCommented Mar 16, 2019 at 6:37
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$\begingroup$ @cherry $v$ is in $V$, which is open. Thus there is a neighbourhood of $v$ which is disjoint from $U$, so $v \notin \overline{U}$. $\endgroup$ Commented Mar 17, 2019 at 17:05
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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.
answered Mar 15, 2019 at 3:39
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$\begingroup$ 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. $\endgroup$– KConradCommented Mar 15, 2019 at 10:01
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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. $\endgroup$ Commented Mar 15, 2019 at 14:57
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1$\begingroup$ Oh, so "$s \in S$" should be "$s \in \overline{S}$". Then my objection no longer applies. $\endgroup$– KConradCommented Mar 15, 2019 at 15:07
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$\begingroup$ @KConrad, Note that $|x-a|=R$ for any $x\in \bar S\backslash S$, therefore $s\in S$. $\endgroup$ Commented Mar 16, 2019 at 0:04