# Grothendieck group of integer sets with Minkowski addition

Consider a set $Z := \{X\in 2^\mathbb{Z}: \Vert X\Vert < \infty\}\setminus\{\emptyset\} = \{X\in 2^\mathbb{Z}: \Vert X\Vert\in \mathbb{N}_+\}$ and give it an operation of Minkowski addition, that is: $$A + B = \{a + b: a\in A \ \wedge \ b\in B\}.$$ This structure is associative, commutative and has an identity element $\{0\}$. If I'm not mistaken, then it's an abelian monoid. What I'm interested in, is its Grothendieck group. Let us have a(n) (equivalence) relation $\sim$ on $Z\times Z$ defined as: $$(A, B)\sim (C, D)\iff A+D+K = C + B + K$$ for some $K\in Z$.

My question is: What is the structure of the group $\ Z\times Z/\sim\$?
Is it described in more detail somewhere?

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Possible hints for me:

There seems to be no set $K$ that would make, for example, $(\{1,2\}, \{2,3\})$ and $(\{-1,1\}, \{3,7\})$ equivalent. If I hadn't excluded infinite sets and the empty one, I could now use $\mathbb{Z}$ (or $\emptyset$, if adding it was defined the way Minkowski did) for $K$ and thus obtain a trivial group.

Now it seems that there's no set $Y\in Z$ such that for any $X\in Z$, $\ Y+X = Y$.
So, first hint I'd appreciate, is to show me, if there really is no such $K$ that makes following statement true: $$\{4, 5, 8, 9\} + K = \{1, 2, 3, 4\} + K.$$ Or perhaps there is one...?
Probably in other words: does the cancellation law hold for this structure?

Secondly, the set of singletons $\{X\in 2^\mathbb{Z}: \Vert X\Vert = 1\} < Z$ is isomorphic with $\mathbb{Z}$. I'd say the same holds for $\{[(\{x\}, \{0\})]: x\in\mathbb{Z} \} < \ Z\times Z/\sim\$, meaning the group really can't be trivial.

On the other hand, $(\{1,2\}, \{2,3\}) \sim (\{3,4\}, \{4,5\})$ for $K = \{0\}$, so it's not isomorphic to $Z\times Z$ either (i.e. $\sim$ isn't an identity relation). I imagine a hint would be to show me how to think about, for example, $[\{0\}, \{1,2\}]$ ("negative $\{1,2\}$ set").

The Grothendieck group of $$Z$$ is the free abelian group on 2 generators $$\mathbb{Z}^2$$.
To see this, first consider the equality: $$A + \mathrm{Conv}(A) = \mathrm{Conv}(A)+\mathrm{Conv}(A)$$ with $$\mathrm{Conv}(A)$$ the interval defined by $$\mathrm{Conv}(A) = \{ a_-,~ a_- +1,\dots,~ a_+ \}$$ with $$a_-$$ and $$a_+$$ the minimal resp. maximal element of $$A$$.
The equality above shows that $$Z$$ is not cancellative. The elements $$A$$ and $$\mathrm{Conv}(A)$$ determine the same elements in the Grothendieck group. So the Grothendieck group of $$Z$$ can be computed as the Grothendieck group of the submonoid $$Z' \subseteq Z$$ consisting of the intervals $$[a,b] = \{a,~a+1,\dots,b\}$$ for $$a,b \in \mathbb{Z}$$, $$a \leq b$$. It is easy to check that the Minkowski sum of two intervals $$[a,b]$$ and $$[c,d]$$ is given by $$[a+c,b+d]$$. This means that the map $$Z' \longrightarrow \mathbb{Z}\times\mathbb{N},\qquad [a,b] \mapsto (a,b-a)$$ is an isomorphism of monoids.
So the Grothendieck group of $$Z$$ agrees with the Grothendieck group of $$\mathbb{Z}\times\mathbb{N}$$, which is $$\mathbb{Z}^2$$.