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Let $\Omega_1 \supset \Omega_2 \supset ...$ a sequence of nonempty, open, bounded and convex sets. Define $\Omega = \operatorname{int} \Bigl( \overline{\displaystyle\bigcap_{k=1}^{\infty} \Omega_k } \ \Bigl) $ and supoose that $\Omega$ is nonempty, open, convex and bounded. My intuition says that $\partial \Omega_k \rightarrow \partial \Omega$ in the Hausdorff distance. I don't know how to prove this. If I prove this then I understand a theorem. Can someone give me a hint to prove this?

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  • $\begingroup$ Don't you want to intersect the closures of the $\Omega_k$? If you intersect the open sets and then take the closure, you might get an empty intersection. Anyway, even if you intersect the closures, you might end up with a set without interior, so you maybe should assume that $\Omega$ is non-empty. $\endgroup$ – Lukas Geyer Dec 5 '13 at 13:59
  • $\begingroup$ Can you assume that $\Omega$ is non empty? Otherwise it is not clear what is the Hausdorff distance from an empty set... $\endgroup$ – Emanuele Paolini Dec 5 '13 at 14:24
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    $\begingroup$ If you can read a bit in French, I recommend springer.com/mathematics/book/978-3-540-26211-4 In chapter 2 there is a detailed discussion of different types of convergence of sets. There is an exercice which resembles your question. If you prove that $\Omega_n$ converges to $\Omega$ in the Hausdorff distance, then the convexity should imply the convergence of the boundaries. $\endgroup$ – Beni Bogosel Sep 15 '14 at 20:23
  • $\begingroup$ Thanks for comment in my old question =) .Thanks for the suggestion. I give to you +1 =D $\endgroup$ – math student Sep 16 '14 at 2:50
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Here is a rough idea: The limit of the boundaries always contains the boundary of the limit, even without convexity. If there is a limit point $x$ of the boundaries of the $\Omega_k$ that is not in $\partial \Omega$, then you can pick another point $y \in \Omega$. If $L$ is the line segment between $x$ and $y$, convexity implies that there are line segments $L_k \subseteq \Omega_k$ such that $L_k \to L$ in Hausdorff distance. Since $x$ is not in the closure of $\Omega$, there is a small disk $D$ about it which is disjoint from $\Omega$. So $\Omega_k \cap D$ has to be contained in a $\epsilon_k$-neighborhood of $L \cap D$, with $\epsilon_k \to 0$. This shows that any line segment from $x$ to a point in $\Omega_k \setminus D$ has to have a slope which is in a $\delta_k$-neighborhood of the slope of $L$, with $\delta_k \to 0$. Passing to the limit, this shows that any line segment in $\Omega$ is parallel to $L$, so that $\Omega$ is contained in a line, contradicting openness.

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  • $\begingroup$ Hi @LukasGeyer . Do you know a book (or article) with this result ? I am interessed in this type of results $\endgroup$ – math student Apr 8 '14 at 21:08
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That's true. You could prove that the sequence $\bar \Omega_k$ converges to $\bar \Omega$ in the Hausdorff distance. In fact $\bar \Omega_k$ being bounded have a subsequence converging somewhere. But being decreasing, the whole sequence converges, and the limit is $\bar\Omega$.

Now you want to prove that $d(\partial \Omega,\partial \Omega_k) \le d(\bar \Omega,\bar \Omega_k)$. To achieve this the nontrivial part is to prove that given $p\in\partial \Omega$ then there exists $q\in \partial \Omega_k$ which is close to $p$. Since $\Omega$ is convex you can find an half line external to $\Omega$ with end-point in $p$. On this line there is some $q\in \partial \Omega_k$. Notice that $d(p,q) = d(q,\bar\Omega) \le d(\bar\Omega_k,\bar\Omega)$.

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  • $\begingroup$ Where are you using convexity in your argument? Without convexity the statement is false. $\endgroup$ – Lukas Geyer Dec 5 '13 at 14:30
  • $\begingroup$ I added some explanation. Honestly I didn't check all the details... but I think the proof could be completed. $\endgroup$ – Emanuele Paolini Dec 5 '13 at 14:39

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