I have a sequence of domains in $R^n$ , $\Omega_1 \supset \Omega_2 \supset...$ where $\Omega_k$ is open, convex and bounded for each $k.$ Define $\tilde{\Omega} = \bigcap_{k=1}^{ \infty} \Omega_k$ and suppose that $\tilde{\Omega} \neq \emptyset$ and suppose that $\tilde{\Omega}$ is open and convex. Is it true that $\mathrm{int}(\mathrm{Cl} ({\tilde{\Omega} })) \subset \tilde{\Omega}$ ? (int denotes the interior of a set and Cl the closure)

If this fact is true, then I understand the proof of a theorem that I am studying.

Someone can help me to prove or disprove the fact ? I have no idea how to start. Thanks in advance

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    $\begingroup$ The $\Omega_k$ are subsets of a finite-dimensional space, $\mathbb{R}^n$ or $\mathbb{C}^n$? $\endgroup$ – Daniel Fischer Sep 8 '13 at 0:00
  • $\begingroup$ the domains are in $R^n$. i fixed my error. thanks for see my error (my english is terrible, sorry ..) $\endgroup$ – math student Sep 8 '13 at 0:04

It is true. $\tilde{\Omega}$ is convex, as the intersection of convex sets.

Let $U = \operatorname{int}\left(\overline{\tilde{\Omega}}\right)$. If $U = \varnothing$, there is nothing to show, so let's assume that $U \neq \varnothing$, and let $x \in U$ arbitrary.

Since $U$ is open, there is an $\varepsilon > 0$ with $B_{2\varepsilon}(x) \subset U$. Let $\delta = 2^{-n-4}\varepsilon$. Since $\tilde{\Omega}$ is dense in $\overline{\tilde{\Omega}}$, hence a fortiori in $U$, there are points $x_i^+ \in B_\delta(x + \varepsilon\cdot e_i) \cap \tilde{\Omega}$ and $x_i^- \in B_\delta(x - \varepsilon\cdot e_i) \cap \tilde{\Omega}$ ($e_i$ is the $i$-th standard basis vector).

Since $\tilde{\Omega}$ is convex, the convex hull of $\{ x_i^+ : 1 \leqslant i \leqslant n\} \cup \{ x_i^- : 1 \leqslant i \leqslant n\}$ is contained in $\tilde{\Omega}$. But that convex hull contains $x$ in its interior, hence $x \in \tilde{\Omega}$. $x \in U$ was arbitrary, hence $U \subset \tilde{\Omega}$.

  • $\begingroup$ beatiful solution! $\endgroup$ – math student Sep 8 '13 at 1:00

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