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Let $\Omega_1 \supset \Omega_2 \supset \dots$ be a sequence of bounded, open and convex sets in $R^n$ with $\Omega :=\operatorname{int}(\overline{\bigcap \Omega_n})$ nonempty. It seems that $\partial \Omega = \partial(\operatorname{int}(\bigcap \Omega_n)) $. But I don't know how to prove or disprove this. Someone could give me a help?

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  • $\begingroup$ What space are you working in? $\endgroup$ – Rob Arthan Jul 14 '15 at 23:40
  • $\begingroup$ in $R^n$ . I forget to write this. sorry. thanks for your question $\endgroup$ – math student Jul 15 '15 at 1:33
  • $\begingroup$ Ok, so here's a hint: Without loss of generality you can assume that $0 \in \Omega$. Now think about how $\Omega$ and $\partial\Omega$ meet each line through $0$. $\endgroup$ – Rob Arthan Jul 15 '15 at 2:06
  • $\begingroup$ Unfortunately I am not seeing how to use this. please could you give me more hints ? $\endgroup$ – math student Jul 15 '15 at 3:48
  • $\begingroup$ It is called "tipless cone", see math.stackexchange.com/questions/3260182 (found by Approach0 search engine) $\endgroup$ – Wei Zhong Jun 19 at 9:12
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The intersection of convex sets is convex. Let $C=\bigcup \Omega_n$; this is a convex set.

For a convex set $C\subset\mathbb{R}^n$, one of two possibilities holds

  1. $C$ has empty interior. Then it is contained in a hyperplane. Consequently, the closure of $C$ has empty interior. You have assumed this doesn't happen.

  2. $C$ has nonempty interior. Then it contains some open ball $B$. For every point $x\in \partial C$ the "cone" $$C(x) = \{tx+(1-t)y:y\in B, t\in[0,1]\}$$ is an open set contained in $C$. Note that $x$ is a limit point of $C(x)$. On the other hand, $x$ is also a limit point of an open half-space that lies in the exterior of $C$, just by convexity of $C$. Conclusion: every boundary point of $C$ is the limit point of the interior of $C$ and of the exterior of $C$. For a set with this property, taking the closure or interior will not change its boundary.

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