# Requirement for connected sets [on hold]

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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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 $$\overline{B(a,s)} = B'(a,s)$$.

• 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. – cherry Mar 16 at 6:37
• @cherry $v$ is in $V$, which is open. Thus there is a neighbourhood of $v$ which is disjoint from $U$, so $v \notin \overline{U}$. – Robert Israel 2 days ago

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.

• 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. – KConrad Mar 15 at 10:01
• @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. – Christian Remling Mar 15 at 14:57
• Oh, so "$s \in S$" should be "$s \in \overline{S}$". Then my objection no longer applies. – KConrad Mar 15 at 15:07
• @KConrad, Note that $|x-a|=R$ for any $x\in \bar S\backslash S$, therefore $s\in S$. – Anton Petrunin Mar 16 at 0:04