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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

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  • $\begingroup$ where you have the closure in each case it changes, should it be the same in each case? $\endgroup$ – Ellya May 7 '14 at 19:22
  • $\begingroup$ the closures are in the manner that i writed =) $\endgroup$ – math student May 7 '14 at 19:36
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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 $$ \partial \left(\bigcap \overline{\Omega_n}\right) = \{ t\xi: t = \rho (\xi) \} = \partial\Omega $$

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