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Suppose $A \subseteq \mathbb N^+$ is any arbitrary set. I want to determine under which conditions some finite set of translates $\{A + k_i : 1 \leq i \leq n\}$ of $A$ cover almost all of $\mathbb N^+$, i.e. when there is a finite set of natural numbers $\{k_i : 1 \leq i \leq n\}$ such that $$\mathbb N^+ \setminus \bigcup_{i=1}^n (A + k_i)$$ is finite.

It seems to me that a natural indicator of this possibility is the natural density of $A$ being positive; but perhaps it also suffices that the upper density is positive, or that the lower density is positive.

Question: Is it the case that a finite set of translates of $A$ covers almost all $\mathbb N^+$ if $A$ has positive upper density? If not, does positive lower density suffice?

feralin
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1 Answers1

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Let $A$ be the set of numbers that are not of the from $m!+r$ with $0\le r<m$. Then $A$ has (lower/upper/normal) density $1$, but more importantly, it has arbitrarily large gaps. Hence $\bigcup (A+k_i)$ will not completely cover any gap of size $>\max\{k_1,\ldots, k_n\}$.

This example ledas us to

Theroem. $A$ has the "finite-translation-cover property" if and only if the gaps in $A$ are bounded.

Proof. If the gaps are not bounded, the the same argument as for the concrete example applies. On the other hand, if all gaps are $\le N$, then the translates by $1,2,\ldots,N$ cover almost all of $\Bbb N$. $\square$

Hagen von Eitzen
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  • I see. So a counterexample would be any set $A$ which is the complement of a set which 1) has upper density less than 1, and 2) has arbitrarily long contiguous subsets. – feralin May 19 '17 at 17:00
  • Before I accept the answer, could you please offer a proof that unbounded gaps implies no finite translation-cover? It sounds intuitively right, but I want to see a more rigorous proof. – feralin May 19 '17 at 17:03