First of all let me define what is the the complementary Hausdorff distance between two open sets, I denoted by $d^{H}$ the usual Hausdorff distance in $\mathbb{R}^n$.

Let $\Omega_1$ and $\Omega_2$ two open subsets of a (large) compact set $B \subset \mathbb{R}^n$, then their complementary Hausdorff distance is defined by : $$d_{H}(\Omega_1 , \Omega _2) := d^{H}(B \setminus \Omega_1 , B\setminus \Omega_2)$$

Let $\Omega_n$ be bounded open subsets of $\mathbb{R^n}$ such that $\Omega_n$ are convex and converge to a nonempty convex open set $\Omega$, in the sense of Hausdorff complementary metric.

I would like to prove that the closure of the sequence $\Omega_n$, denoted $\overline{\Omega_n}$, converges to $\overline{\Omega}$ in the usual Hausdorff metric.

In other words :

($\Omega_n \longrightarrow \Omega $ in the complementary Hausdorff metric )$\Longrightarrow( \overline{\Omega_n} \longrightarrow \overline{\Omega}$ in the usual Hausdorff metric ).

  • $\begingroup$ Why do you say "if" after "In other words"? $\endgroup$ – mathworker21 Apr 30 '18 at 4:15
  • $\begingroup$ it's just a topos , thank you $\endgroup$ – Bernstein Apr 30 '18 at 9:40

Assume that $B$ is a large closed ball containing all $\Omega_n$.

(1) Since $B-\Omega_n \rightarrow B-\Omega$, if $B(p,r)$ does not intersect $ B-\Omega$, then $B(p,r/2)$ does not intersect $B-\Omega_n$.

That is, $p\in \Omega$ implies $p\in \Omega_n $.

Blaschke's Theorem implies that if $ \overline{\Omega}_n $ has a limit $C$, then $p\in C$ so that $\Omega \subset C$.

(2) If $\overline{\Omega}=C$, then we are done.

If not, then there is a point $q\in C$ s.t. open ball $B(q,R)$ does not intersect $ \Omega$.

Since $\Omega$ is open, so it contains some open ball. Here from convexity of $C$, we have $$C\ \bigcap\ \overline{B(q,R/2)},$$ which contains some closed $\delta$-ball $B_1$.

If $S_n$ is $\varepsilon_n$-net for $B-\Omega_n$, then $$d_H(S_n,B-\Omega )\leq d_H(S_n,B-\Omega_n) + d_H(B-\Omega_n,B-\Omega)<\varepsilon_n+\epsilon_n <\delta$$

That is, there is $z_n\in S_n\ \cap\ B_1$.

Since $B_1$ is compact, then $z_n\rightarrow z\in B_1$. And $ z_n\in B-\Omega_n \rightarrow z\in B-\Omega$

It is a contradiction.

  • $\begingroup$ Dear @HK Lee thank you for the interest , i have a problem what this mean : $S_n$ is $\varepsilon_n -$ net for $B-\Omega_n$ ? $\endgroup$ – Bernstein Apr 30 '18 at 12:31
  • $\begingroup$ A subset $N$ of a metric space $(X,d_X)$ is $\epsilon$-net if any element $x$ in $X$ has $d_X(x,N)\leq \epsilon$. $\endgroup$ – HK Lee Apr 30 '18 at 12:36
  • $\begingroup$ Thank you @HK Lee,I am too excited, because I have this conjecture in mind and I believe that it is true your demontration. could better explain to me about the contradiction why ? $\endgroup$ – Bernstein Apr 30 '18 at 12:50
  • $\begingroup$ very nice proof , can i ask you one question what is your level in math? can we have a private discussion ? I do not know if this site offers this $\endgroup$ – Bernstein Apr 30 '18 at 13:22
  • $\begingroup$ Yes. I can have a discussion. I am a basic level. I read a part of a convex analysis book and a part of undergraduate metric geometry book. $\endgroup$ – HK Lee Apr 30 '18 at 13:35

May assume that the origin $0$ is inside $\Omega$.

Denote by $d$ the usual distance between sets ($\inf$ of pairwise distances).

  1. For any $0<\epsilon <1$ we have $$d((1-\epsilon)\Omega, \Omega^c)= \delta_{\epsilon}>0$$ so $\Omega_n^c$ will have no points in $(1-\epsilon)\Omega$ for $n$ large enough, that is $$\Omega_n \supset (1-\epsilon)\Omega$$ for $n>n_{\epsilon}$

Let $n> n_{1/2}$, so that $\Omega_n$ contains $\frac{1}{2}\Omega$. Let $B$ be a ball of radius $r$ with center $0$ inside $\frac{1}{2}\Omega$. Assume that $\Omega_n$ contains some point $x$ outside of $\Omega$. We'll show that point has to be pretty close to $\Omega$ for $n$ large. Let $x= (1+\epsilon) x_0$, where $x_0$ is on the boundary of $\Omega$. Now $\Omega_n$ contains $x$ and $B$ so it will contain the convex hull $C$ of $\{x\}\cup B$. The distance from $x_0$ to the complement of $C$ is $\frac{\epsilon}{1+\epsilon} r$ (make a picture in 2D). Therefore, for $n>n'_{\epsilon}$ we have $$\Omega_n\subset (1+\epsilon)\Omega$$.

Now we only have to check that as $\epsilon \to 0$ $$(1\pm \epsilon) \bar \Omega\to \bar\Omega$$ in the usual Haausdorff metric (use compactness of $\bar \Omega$).

  • $\begingroup$ Thank you for you interest @orangeskid , the condition thats $\Omega_n^c$ have no point with $(1-\epsilon)\Omega $ i think if the sequences of sets $\Omega_n$ is increasing. this is not true , what do you think ? $\endgroup$ – Bernstein Apr 23 '18 at 11:59
  • $\begingroup$ $\Omega_n^c$ is the complement of $\Omega_n$. If any point from $\Omega_n^c$ lies inside $(1-\epsilon)\Omega$ then its distance to $\Omega^c$ is $\ge \delta_\epsilon$. This can happen only for finitely many $n$. Best to draw a picture. $\endgroup$ – Orest Bucicovschi Apr 23 '18 at 12:19
  • $\begingroup$ yes this what i mean , so your prrof is based on this for$ n\geq n_0 $ , you want to do this $(1+\epsilon)\Omega \subset \Omega_n\subset (1+\epsilon)\Omega$ , so you take $\epsilon \rightarrow 0 $ ,one gets $(1\pm \epsilon) \bar \Omega\to \bar\Omega $and where do you use the complementary Haussdorff distance and the consclusion ? $\endgroup$ – Bernstein Apr 23 '18 at 12:33
  • $\begingroup$ @Bernstein: it's used twice in the proof. need to go carefully through it. $\endgroup$ – Orest Bucicovschi Apr 23 '18 at 12:50
  • $\begingroup$ excuse me , maybe I do not see the links between $ (1-\epsilon)\Omega \subset \Omega_n\subset (1+\epsilon)\Omega$ and the conclusion could you write me more From this inclusions to the conclusiosn, $\endgroup$ – Bernstein Apr 23 '18 at 13:13

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