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I am trying to prove a fact that is easy to see, but i dont know how to prove...

I believe that is true. The fact is :

Consider $\Omega$ a convex , open , bounded domain in $R^n$.Let $\tilde{\Omega} \subset \Omega $ a open set. If $\partial \Omega = \partial \tilde{\Omega}$, then $\tilde{\Omega} = \Omega$.

Someone can give me a hint to prove or disprove the affirmation ?

Thanks in advance

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  • $\begingroup$ is trivial prove this ? $\endgroup$ – math student Sep 24 '13 at 12:52
  • $\begingroup$ Pretty much yes. It may not be trivial to see the trivial proof (probably isn't), but there's not much involved. $\endgroup$ – Daniel Fischer Sep 24 '13 at 12:58
  • $\begingroup$ @DanielFischer : could you please give some hint to follow up... Do we have to take $v\in V $ and prove that $v\in U$.. AS $V\cap \partial V =\emptyset$ I am not able to move further... $\endgroup$ – user87543 Sep 24 '13 at 13:08
  • $\begingroup$ @PraphullaKoushik What relations do you know between $A, \overline{A}, \partial A, \overset{\circ}{A}$? $\endgroup$ – Daniel Fischer Sep 24 '13 at 13:10
  • $\begingroup$ In general $\bar{A}=A\cup \partial A$ $\endgroup$ – user87543 Sep 24 '13 at 13:12
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In any topological space $X$, for any subset $A \subset X$, we have the disjoint union

$$X = \overset{\circ}{A} \,\dot{\cup}\, \partial A\,\dot{\cup}\, (X\setminus A)^\circ.$$

In particular, we have $\overline{A} = A \cup \partial A = \overset{\circ}{A}\,\dot{\cup}\, \partial A$.

For an open $V \subset X$, we have $\overset{\circ}{V} = V$, so $\overline{V} = V \,\dot{\cup}\,\partial V$ and $X = V \,\dot{\cup}\,\partial V\,\dot{\cup}\, (X\setminus V)^\circ$.

If we also have an open $U \subset V$ with $\partial U = \partial V$, then we have the decomposition

$$V = U \,\dot{\cup}\, \bigl(V\cap (X\setminus U)^\circ\bigr)$$

of $V$ into disjoint open sets. If $V$ is connected, one of the two must be empty, so $U \neq \varnothing$ implies $U = V$, and $U = \varnothing$ implies $\partial V = \varnothing$ (which in general does not imply $V = \varnothing$, but under the given assumptions does).

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  • $\begingroup$ thanks for your help!. Do you know a counter example if V is not open ? $\endgroup$ – math student Sep 24 '13 at 23:08
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    $\begingroup$ Take a closed ball in $\mathbb{R}^n$, and the corresponding open ball. Or two sets between the open and the closed ball. $\endgroup$ – Daniel Fischer Sep 24 '13 at 23:54
  • $\begingroup$ I did a mistake. its U not open. sorry... $\endgroup$ – math student Sep 25 '13 at 0:22
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    $\begingroup$ It's not possible in locally connected spaces. In a locally connected space, if $V$ is open, and $U \subset V$ with $\partial U \subset \partial V$, then $U$ is a union of connected components of $V$, hence open. It's also not possible if $V$ is connected, for the same reason, if $U \cap C \neq \varnothing$ and $C\setminus U \neq \varnothing$, where $C$ is a connected component of $V$, then $\partial U \cap C \neq \varnothing$, otherwise you'd have a decomposition of $C$ into the parts interior and exterior to $U$. $\endgroup$ – Daniel Fischer Sep 25 '13 at 19:48
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    $\begingroup$ It could be that one can construct an example with a disconnected open set $V$ in a not locally connected space. But I don't know whether it's possible. $\endgroup$ – Daniel Fischer Sep 25 '13 at 19:50

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