Is there a normed vector space of shapes? I've recently been interested in the Hausdorff Distance (a notion of distance between sets/shapes), and I am curious if there is a normed (or semi-normed) vector space of sets in $\mathbb{R}^n$, where the Hausdorff distance induces the norm or seminorm. Restricting to a subset of "sets in $\mathbb{R}^n$" (e.g., compact, convex sets) would be fine. However, let's exclude the "cheeky" answer which is just taking the collection of "all singletons in $\mathbb{R}^n$".
Defining an algebraic notion of addition is the biggest question I'm stuck on. Minkowski addition seems like a natural candidate, although additive inverses do not usually exist for Minkowski addition.
EDIT: Here is a construction I'm not sure about:
Let $M$ be the collection of all convex sets in $\mathbb{R}^n$ under Minkowski addition; then we mod out by the equivalence relation where all balanced sets are the same. Under this equivalence relation, we would at least get additive inverses, because for $C\in M$, $C-C$ is always balanced in $\mathbb{R}^n$, i.e., $C-C\equiv \{0\}$ under this relation. I imagine this probably does not work out with the Hausdorff metric though, since this equivalence relation will likely distort distances.
 A: Yes, you are right on the money here! Given a normed linear space $X$ (not necessarily finite-dimensional), the set of closed, bounded, convex, non-empty subsets of $X$ (let $\mathcal{C}(X)$ be the set of such sets) can be embedded into a larger normed linear space, where the norm is an extension of the Hausdorff distance, and the addition is the closure of the Minkowski sum:
$$A \oplus B := \overline{A + B} = \operatorname{cl}\{a + b \in X : a \in A, b \in B\}.$$
Scalar multiplication is defined as usual:
$$\lambda A = \{\lambda a : a \in A\},$$
where $\lambda$ is non-negative (the negative case has to be handled a little differently).
I actually did my honours thesis on this space, which culminated in a paper that I co-wrote with my honours supervisor.
Unfortunately, it's not quite so simple as saying these convex sets form a vector space; clearly the set $\{0\}$ would be the identity of $\oplus$, but there can be no additive inverse to a set $C$ unless $C$ is a singleton (in which case $\{x\}$ and $\{-x\}$ are inverses).
So, instead, we embed them in a larger space $\mathcal{R}(X)$. We create this larger space in a similar way to how we can define $\Bbb{Z}$ in terms of $\Bbb{N}$, or how we define $\Bbb{Q}$ in terms of $\Bbb{Z}$. What we do is create an equivalence relation on $\mathcal{C}(X) \times \mathcal{C}(X)$:
$$(A, B) \sim (C, D) \iff A \oplus D = B \oplus C.$$
The equivalence classes become our space. In the paper, we denote the equivalence class by $A \ominus B$.
The sets $A \in \mathcal{C}(X)$ become identified with the equivalence class containing $(A, \{0\})$, i.e. $A \ominus \{0\}$. We can extend $\oplus$ to the entire space by defining:
$$(A \ominus B) \oplus (C \ominus D) = (A \oplus C) \ominus (B \oplus D).$$
Scalar multiplication is defined by:
$$\lambda(A \ominus B) = \begin{cases} \lambda A \ominus \lambda B & \text{if } \lambda \ge 0 \\ (-\lambda)B \ominus (-\lambda)A & \text{if } \lambda < 0.\end{cases}$$
Under these operations, the space $\mathcal{R}(X)$ is a vector space. This vector space turns out to be a Kakutani space, which I won't go into, but has a very natural choice of norm, which comes to, in this case:
$$\|A \ominus B\| = d_H(A, B),$$
i.e. the Hausdorff distance between the two sets!
And yes, these operations, and norm, turn out to be well-defined over this equivalence relation.
So, yes, well spotted! We can make a vector space out of sets, using Minkowski addition and the Hausdorff metric.
