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Let $E \subset R^n$, where $E^c$ is disconnected.

Then exists $U,V \subset R^n$, $U,V \neq \emptyset$ disjoint, and open relative to $E^c$ , with $E^c = U \cup V.$

We have too $U \cap \overline{V} = \emptyset = V \cap \overline{U} $ . My book says :

Exists a ball $B \subset R^n$ such that $B$ intersects both ${\overline{U}}^c$ and ${\overline{V}}^c$.

Drawing a picture is easy to see this last affirmation, but i dont know how to prove that. Someone can give a hint ? Any help is apreciated.

Thanks in advance!

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  • $\begingroup$ Are you sure this is the question you wanted to ask? Because every point belongs either to $U$ or $V$ but not both, so any ball that contains at least one point from $U$ and one point from $V$ satisfies your requirement. This is quite trivial. Also, I don't see how $E$ is relevant to the question, so why do you mention it at all? $\endgroup$ – Marek Aug 12 '13 at 17:39
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Take points $x\in U$ and $y \in V$. Then a ball of radius $>d(x,y)$ around either should work.

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$\overline U^c$ and $\overline V^c$ are open so if this ball don't exist, we have: (1) $\overline U^c \cap \overline V^c = \emptyset$ hence (deMorgan's law) $\overline U \cup \overline V = \mathbb R = \overline{U \cup V}$. So $U \cup V$ is dense in $\mathbb R$. But we have $\overline V \subset \overline U^c$ and $\overline U \subset \overline V^c$. (from condtitons between $U$, $\overline U$ , $V$ and $\overline V$)

$\overline U^c$ and $ \overline V^c$ are open and (1) - contradiction.

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