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

Question

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 ).

Answer 1

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.

Answer 2

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 $\bar \Omega$).