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Consider $$\Omega_1 \supset \Omega_2 \supset \cdots$$ a sequence of bounded, open and convex domains in $\mathbb R^n$, with all the inclusions strict. I want to prove that $\bigcap \Omega_k \subset \overline{\bigcap \Omega_k} \subset \Omega_k$ for all $k$. I believe this is true. In some cases is easy to see this. but i dont know how to prove . Someone can give me a hint to how to prove or disprove ? the problem is the second inclusion

Thanks in advance

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  • $\begingroup$ By "$\cap\Omega_k$ do you mean $\bigcap_k\Omega_k$? $\endgroup$ – Michael Hardy Sep 13 '13 at 2:23
  • $\begingroup$ @MichaelHardy , yes =) $\endgroup$ – math student Sep 13 '13 at 2:24
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The second inclusion is sometimes false. E.g. in $\mathbb R$, $\Omega_k=(0,1+1/k)$.

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As for the first inclusion, it is obviously weakly true, since every set is included in its closure by definition. But the inclusion need not be strict. Let, for each $k\in\mathbb{Z}_+$, $\Omega_k\equiv(-1/k,1/k)$. Then, the intersection $\bigcap_{k\in\mathbb{Z}_+}\Omega_k$ is $\{0\}$, which is a closed set, so that it is equal to its closure.

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