a decreasing sequence of convex domains Let $\Omega_1 \supset \Omega_n \supset\cdots$ a decreasing sequence of bounded, convex and open sets in $R^n$. Define $\Omega = \operatorname{int} \left(\overline{\bigcap \Omega_n}\right)$ and supose that $\Omega \neq \emptyset$ , convex , open and bounded. Then is true that $\partial \left(\bigcap \overline{\Omega_n}\right) = \partial \Omega$?
Intuitively this is true, but i dont know if is true. If is true it will help me understand the proof of a theorem. 
To prove i am trying to use this : $\operatorname{int}(\overline{A}) = \overline{A}$ for every bounded convex set in $R^n.$
someone can give me a help to prove or disprove what i am asking? 
thanks in advance
 A: Yes, this is true. Recall that a convex set has nonempty interior if and only if it  is  not contained in any hyperplane. Since  the closure of $\bigcap \Omega_n$ has nonempty interior, it is not contained in a hyperplane. Hence, $\bigcap \Omega_n$ is not contained in a hyperplane. So it has an interior point. We may assume it's the origin; so, $0\in \Omega_n$ for all $n$. 
By convexity, every $\Omega_n$ can be described as $\Omega_n = \{ t\xi: 0\le t<\rho_n(\xi)  \} $ where $\xi$ runs over all unit vectors. Let $\rho(\xi) = \inf_n \rho_n(\xi)$. It is routine to verify that: 
$$
\{ t\xi: 0\le t<\rho (\xi)  \} \subseteq \bigcap_n \Omega_n \subseteq \{ t\xi: 0\le t\le \rho (\xi)  \}
$$
hence, 
$$ \overline{\bigcap \Omega_n}  = \{ t\xi: 0\le t\le \rho (\xi)  \}$$
and 
$$\Omega =  \{ t\xi: 0\le t<\rho (\xi)  \},\quad 
\partial \Omega = \{ t\xi:  t = \rho (\xi)  \}$$
Along the same lines, 
$$
 \bigcap \overline{\Omega_n}  = \{ t\xi: 0\le t\le \rho (\xi)  \}
$$
which has the boundary
$$
 \partial \left(\bigcap \overline{\Omega_n}\right)  = \{ t\xi: t = \rho (\xi)  \} = \partial\Omega
$$
