# The boundary of the intersection of a decreasing sequence of convex sets

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?

• What space are you working in? – Rob Arthan Jul 14 '15 at 23:40
• in $R^n$ . I forget to write this. sorry. thanks for your question – math student Jul 15 '15 at 1:33
• 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 $0$. – Rob Arthan Jul 15 '15 at 2:06
• Unfortunately I am not seeing how to use this. please could you give me more hints ? – math student Jul 15 '15 at 3:48
• It is called "tipless cone", see math.stackexchange.com/questions/3260182 (found by Approach0 search engine) – Wei Zhong Jun 19 at 9:12

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.