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Given $n$ distinct positive real numbers $A = \{a_1 \ldots a_n\}$, I would like to find a positive lower bound for the absolute value of the non-zero elements of the $\mathbb{Z}$-span $$\langle A\rangle_{\mathbb{Z}} := \mathbb{Z} a_1 + \ldots + \mathbb{Z} a_n\ .$$ In fact, all I need is that there is a positive lower bound on the absolute value, i.e. some $d > 0$ such that $|x| \geq d$ for all $x \in \langle A\rangle_{\mathbb{Z}}\setminus\{0\}$. I don't care what $d$ looks like exactly.

What is the easiest way to prove this? I tried a straightforward induction on $n$ but did not get very far.

(This isn't exactly a lattice, as far as I know, because the generators are not linearly independent, but it seems sufficiently similar, so I tagged the question with ‘lattices’)

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    $\begingroup$ There are essentially just two cases. Case 1: all of the $a_i/a_j\in\mathbb{Q}$. Then it is reduced to the rational (hence integral) case. Case 2: at least one $a_i/a_j\notin\mathbb{Q}$. Then the lattice is dense in $\mathbb{R}$, and the answer is 0. $\endgroup$
    – Fan Zheng
    Commented Nov 9, 2016 at 20:29
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    $\begingroup$ Argh! Yes, I think you're right. It never occurred to me that the problem may be that my claim is quite simply wrong. Thank you! What I actually seem to need is something weaker, namely that any sequence in the $\mathbb{N}$-span of $A$ tends to infinity. Any thoughts on that? $\endgroup$ Commented Nov 9, 2016 at 21:11
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    $\begingroup$ Could you clarify what is meant by " any sequence in the $\mathbb{N}$-span of $A$ tends to infinity"? $\endgroup$
    – Fan Zheng
    Commented Nov 9, 2016 at 22:55
  • $\begingroup$ Since the a_i are positive, the sum of n of them is larger than n times the smallest of them. I am not sure where you are experiencing difficulty. (If you need to have negative terms, then you don't have such an increasing lower bound.). Gerhard "Is This What You Want?" Paseman, 2016.11.09. $\endgroup$ Commented Nov 10, 2016 at 4:02
  • $\begingroup$ Well, you may use every a_i arbitrarily often in the sum. Still, I actually realised yesterday that the problem is trivial, seeing as there can obviously be only finitely many linear combinations that are smaller than any given bound. $\endgroup$ Commented Nov 10, 2016 at 7:31

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