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this seems intuitive to me but I'm struggling to prove it (is it false?).

Let $E$ be a subset of a lattice (free abelian group of finite rank) closed under addition, containing the origin and such that if $e \in E$ then $-e \notin E$ (it's some sort of "positive" cone, oh actually the correct term should probably be submonoid).

If I take any other element $x$ and consider the intersection of $E$ with $-(E +x)$, will it be finite (or empty)?

(All the properties considered are invariant under isomorphism, so we might as well consider the problem in $\mathbb{Z}^n$)

[please help me out for tags, it's not my usual cup of tea]

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@Nunoxic The previous edit was a bit Weird. So, I've rolled it back to the OP's version. –  user21436 Feb 14 '12 at 12:10
    
@KannappanSampath No issues. I just thought the title seemed too big. –  Inquest Feb 14 '12 at 12:11
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1 Answer 1

up vote 2 down vote accepted

Since you require $0\in E$ you must exempt $x=0$ from the requirement that $x\in E$ implies $-x\notin E$. If you do, then $E\cap −(E+x)$ may be infinite: take $E=\{(i,j\in\mathbf Z^2\mid i>0 \lor (i=0 \land j\geq0)\}$, and $x=(-n,0)$ with $n\in\mathbf N_{>0}$.

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silly me! thanks for that. it seems in the situation I have I actually need to do something different. I will probably turn up with another question soon... –  Jacob Bell Feb 15 '12 at 0:37
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