Does convergence of convex sets in Hausdorff distance implies convergence of the complementary sets? Definition:
The Hausdorff distance associated with a distance $d$ on a space $E$ between two sets $A\subset E$ and $B \subset E$ is $d_H(A, B) = \max(\sup_{x\in B}\{d(x, A)\}, \sup_{y\in A}\{d(y, B)\})$, i.e. this is the maximum distance any point of A can be from B and vice versa.
Question:
For a fixed dimension $d$, consider a sequence of convex sets $A_n \subset \mathbb{R}^d$ and a convex set $A \subset \mathbb{R}^d$ such that $d_H(A_n, A) \to 0$ when $n \to \infty$.
Does $d_H(A_n^c, A^c) \to 0$ when $n\to \infty$ ? ($A^c$ denotes the complementary set of $A$)
 A: Yes. For each unit vector $v$ in $\mathbb{R}^d$ and for any set $B \subseteq \mathbb{R}^d$, let $$r(B,v) = \inf_{b \in B} \langle b,v\rangle,$$ where $\langle x, y \rangle$ is the inner product of $x$ and $y$. It is not too hard to show that for any $B,C \subseteq \mathbb{R}^d$, $$|r(B,v) - r(C,v)| \leq d_H(B,C),$$ with the appropriate conventions regarding $\infty$.
First I will argue in the case where the $A_n$'s and $A$ are closed convex sets, and then I will extend this to arbitrary convex sets.
For any closed convex set $B$, we have that $$B = \bigcap_{v\text{ unit vector}} \{x \in \mathbb{R}^d : \langle x, v\rangle \geq r(B,v)\}.$$ Let $H(r,v)$ denote the set $\{x \in \mathbb{R}^d : \langle x, v \rangle < r\}$. We clearly have that $B^c = \bigcup_{v\text{ unit vecotr}}  H(r(B,v),v)$. It's also not too hard to see that for any fixed unit vector $v$, $d_H(H(r,v),H(r',v)) = |r-r'|$.
Now the crux is this: The Hausdorff metric 'commutes' with arbitrary unions in the following sense, if $\{B_i\}_{i \in I}$ and $\{C_i\}_{i \in I}$ are arbitrary families of sets such that $d_H(B_i,C_i) \leq \varepsilon$ for every $i \in I$, then $d_H\left( \bigcup_{i \in I} B_i, \bigcup_{i \in I} C_i) \right) \leq \varepsilon$.
This means that for each $n$, we have that
$$ d_H(A^c_n ,A^c) = d_H\left(\bigcup_{v\text{ unit}}H(r(A_n,v),v), \bigcup_{v\text{ unit}}H(r(A,v),v) \right) \leq \sup_{v\text{ unit}}d_H(H(r(A_n,v),v),H(r(A,v),v)) = \sup_{v\text{ unit}} |r(A_n,v)-r(A,v)| \leq d_H(A_n,A).$$
So we have that $d_H(A_n^c,A^c)$ goes to $0$ as well.
For arbitrary sequences of convex sets, we just need the following facts:

*

*For any set $B$, $d_H(B,\overline{B}) = 0$, where $\overline{B}$ is the closure of $B$.

*For any convex set $B \subseteq \mathbb{R}^d$, $d_H(B^c, \overline{B}^c) = 0$.

The first fact is standard. The second is not too hard but requires a little bit of proof.
