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I believe that I correctly proved the following statement:

Let $\Omega$ be an open, convex domain in $R^n$ . Consider $\tilde{\Omega} \subset \Omega$ a non empty set. Suppose that $\partial \tilde{\Omega} \subset \partial{\Omega}$. Then $ \tilde{\Omega}$ is open with $\tilde{\Omega} = \Omega$.

Proof:

Suppose that $\tilde{\Omega} \neq \Omega$. Then there exists a element $x \in \Omega - \tilde{\Omega}$. We have $\tilde{\Omega} \neq \emptyset$ and $x \notin \tilde{\Omega} $ , then there exists $y \in \tilde{\Omega}$ with $y \neq x$. We have the segment $[y,x] = \{ (1-t)y + t x , 0 \leq t \leq 1 \} \subset \Omega $ (because $\Omega$ is convex). Note that this segment is connected and $[y,x]$ has a point of $\tilde{\Omega}$ and a point of $\Omega - \tilde{\Omega}$. Then there exists $z \in [y,x]$ with $z \in \partial {\tilde{\Omega}} \subset \partial{\Omega}$. Note that $[y,x] \subset \Omega$, then $z \in \Omega$ then we have $z \in \partial \Omega \cap \Omega$. Contradiction, because $\Omega$ is open. Then $\tilde{\Omega} = \Omega$.

Is my proof correct? Any comment will be apreciated =)

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Yes, your proof is correct but with caveats. You should've told that you are using customhouse Theorem.

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    $\begingroup$ What is the "customhouse Theorem"$\hspace{.02 in}$? $\;\;$ $\endgroup$ – user57159 Sep 25 '13 at 0:15
  • $\begingroup$ Consider $X\subset R^n$. If a connected $C\subset R^n$ contains a point $a\in X$ and a point $b\notin X$ then $C$ must contain a point in the boundery of $X$. $\endgroup$ – checkmath Sep 25 '13 at 0:25

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