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We know that any open $ (\mathbb{ R}, d) $ is the union of at most countable open balls disjunct two by two. Show that the same is not true in $ \mathbb{ R}^ n $, if $ n \geq 2 $.

I'm trying to show a counter example where I display an open set of R where A is not a joint of disjoint open balls.

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  • $\begingroup$ The normal English for your disjunct two by two is pairwise disjoint. $\endgroup$
    – Brian M. Scott
    Feb 14, 2021 at 23:02
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    $\begingroup$ You can use any open rectangle, together with the fact that the union of two or more pairwise disjoint open balls is not connected. $\endgroup$
    – Brian M. Scott
    Feb 14, 2021 at 23:04
  • $\begingroup$ I thought to show that there are no open countable and disjunct balls in pairs whose union is the open square $(0,1)^2\subset \mathbb{R}^2$. I know it's because of the singing, but I don't know how to write it in mathematical terms. $\endgroup$
    – User
    Feb 14, 2021 at 23:11
  • $\begingroup$ Use the fact that the open unit square is connected, and prove that if $\mathscr{U}$ is a countable family of pairwise disjoint open balls, then $\bigcup\mathscr{U}$ is not connected. (In fact its components are the members of $\mathscr{U}$.) $\endgroup$
    – Brian M. Scott
    Feb 14, 2021 at 23:15

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