# Convergence of convex sets in the complementary Hausdorff metric and in the usual Hausdorff metric

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 ).

• Why do you say "if" after "In other words"? – mathworker21 Apr 30 '18 at 4:15
• it's just a topos , thank you – 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 $z_n\in B-\Omega_n \rightarrow z\in B-\Omega$

• Dear @HK Lee thank you for the interest , i have a problem what this mean : $S_n$ is $\varepsilon_n -$ net for $B-\Omega_n$ ? – Bernstein Apr 30 '18 at 12:31
• A subset $N$ of a metric space $(X,d_X)$ is $\epsilon$-net if any element $x$ in $X$ has $d_X(x,N)\leq \epsilon$. – HK Lee Apr 30 '18 at 12:36
• 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 ? – Bernstein Apr 30 '18 at 12:50
• 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 – Bernstein Apr 30 '18 at 13:22
• 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. – 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 $\bar \Omega$).

• 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 ? – Bernstein Apr 23 '18 at 11:59
• $\Omega_n^c$ 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. – Orest Bucicovschi Apr 23 '18 at 12:19
• 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 ? – Bernstein Apr 23 '18 at 12:33
• @Bernstein: it's used twice in the proof. need to go carefully through it. – Orest Bucicovschi Apr 23 '18 at 12:50
• 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, – Bernstein Apr 23 '18 at 13:13