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Let $(X,d)$ be a metric space such that for all $x \in X$ and all $r>0$, $\overline{B(x,r)} = \{y \in X \mid d(x,y)\leqslant r\}$ Show that every open ball of $X$ is connected.

Note- I was trying to move with contradiction, but failed!

Daniel Fischer
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UNM
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    It would be interesting to see your proof attempt. You may be on the right track! – Umberto P. Oct 22 '13 at 14:37
  • Which notion of "connected" are you trying to prove it with? – rschwieb Oct 22 '13 at 14:39
  • I agree, especially if it's a homework related question, you should post your attempted efforts as well. – user2566092 Oct 22 '13 at 14:39
  • @Arthur I think you may need something more like $[0,2] \cup [3,5]$ to get a counterexample but yeah it seems like this statement isn't true. – user2566092 Oct 22 '13 at 14:43
  • @user2566092 The closure of the open ball around $2$, of radius $1$ is just $[1,2]$ but the set of points at a distance less than or equal to $1$ from the point $2$ is $[1,2]\cup{3}$. – Dan Rust Oct 22 '13 at 14:45
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    @DanielRust What about $[0,2) \cup (3,5]$ though? – Arthur Oct 22 '13 at 14:48
  • @DanielRust ok fair enough. At any rate, it seems like something is wrong in this problem, unless connected is supposed to mean locally connected or something. – user2566092 Oct 22 '13 at 14:49
  • @Arthur that does seem to be a counterexample... – Dan Rust Oct 22 '13 at 14:51
  • I don't know what's the difference between local connectedness and this one! It's from preliminary parts I can say! – UNM Oct 22 '13 at 14:51
  • My attempt: Let B(x,r) be disconnected! therefore B(x,r)= A U B. where A, B are disjoint, non empty open sets in B(x,r). A is open set hence can be written as union of open balls, then can you find a contradiction? – UNM Oct 22 '13 at 14:54
  • @UNM What is your definition of connectedness? And why is $[0,2) \cup (3,5]$ not a counterexample? – Arthur Oct 22 '13 at 14:55
  • a set A is disconnected if it can be written as union of two non empty separated subsets of A. – UNM Oct 22 '13 at 14:57
  • Well, then it seems that something is wrong with the problem. – Arthur Oct 22 '13 at 14:59
  • Agreed, something is wrong with the problem – user2566092 Oct 22 '13 at 15:00
  • If A is connected subset, is int(A) connected? In general! @Arthur – UNM Oct 22 '13 at 15:09
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    The answer to that is no. Consider the solid balls $B((0,0),1)$ and $B((2,0),1)$ in $\mathbb{R}^2$. Their union is connected but the interior of their union is disconnected. – Dan Rust Oct 22 '13 at 15:17
  • https://math.stackexchange.com/questions/1363810/prove-that-if-the-closure-of-each-open-ball-in-compact-metric-space-is-the-close?rq=1 – Clemens Bartholdy Aug 03 '24 at 09:53

1 Answers1

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The statement is not true. Take for example $X = [0,2) \cup (3,5]$. All properties are satisfied, but the open ball $B(1.5,2) = [0,2) \cup (3,3.5)$ is not connected.

Arthur
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