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

  • $\begingroup$ 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

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).

  • $\begingroup$ thanks for your help!. Do you know a counter example if V is not open ? $\endgroup$ – math student Sep 24 '13 at 23:08
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.