**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 $A$)



Does convergence of convex sets in Hausdorff distance implies convergence of the complementary sets?
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.


