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Please let me know if this proof looks good.

For the sake of contradiction, suppose that $E\ne \emptyset$ and $E\ne \mathbb{R}^n$, and that the boundary of $E$, $\partial E$, is empty. Since the closure of $E$, $\overline{E}$, is the union of the interior of $E$ and the boundary of $E$, i.e. $\overline{E}=E^\circ \cup \partial E$, $\overline{E}=E^\circ$. But $E^\circ$, the interior of $E$, is open by definition. Thus $\overline{E}$ is open. But this is impossible! Therefore, $\partial E$ is not empty.

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  • $\begingroup$ If $\overline E=\Bbb R^n$ then it's both open and closed. $\endgroup$ – Gregory Grant Jan 27 '16 at 23:32
  • $\begingroup$ Yes it's correct. @GregoryGrant: $E$ is a proper ! $\endgroup$ – Surb Jan 27 '16 at 23:33
  • $\begingroup$ @Surb Just because $E$ is proper does not mean $\overline E\not=\Bbb R^n$. $\endgroup$ – Gregory Grant Jan 27 '16 at 23:34
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    $\begingroup$ @Surb sorry I meant $E$ proper does not necessarily imply $\overline E\not=\Bbb R^n$. Like if $E=\Bbb Q^n$ for example. $\endgroup$ – Gregory Grant Jan 27 '16 at 23:36
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    $\begingroup$ But that can easily be handled as a special case. If $\overline E=\Bbb R^n$ then obviously $\overline{(E^c)}\cap\overline E\not=\emptyset$. And no other subset of $\Bbb R^n$ is both open and closed, since $\Bbb R^n$ is connected. $\endgroup$ – Gregory Grant Jan 27 '16 at 23:38
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As others have noted, you have only shown that $\partial E=\phi$ implies that $\bar E$ is open and closed. This means $\partial E=\phi \implies (\bar E=\phi \lor \bar E=R^n)$. Now if $\bar E=R^n$ and $E$ has empty boundary then $\phi= \partial E=\bar E\cap \overline { (R^n\backslash E}=R^n\cap \overline {R^n\backslash E}\supset R^n\backslash E,$ so $E=R^n$. Note that we can apply this result to any connected space, not just $R^n$.

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