The question that I shall ask here has arisen in the context of Diophantine approximation. I find it rather interesting, and I have no idea how to answer it. Any help, advice, or suggestions for references would be much appreciated.

Let $S$ be a countably infinite subset of $\mathbb{R}_{>0}$ that satisfies the following conditions. Let $n$ be an arbitrary element of $\mathbb{Z}_{\geq 2}.$

1) For any integer $k \geq 1,$ we have $k \cdot S \subseteq S.$ (In particular, $S$ is unbounded.)

2) There exists a (necessarily unique) strictly increasing sequence $t_\bullet = (t_k)_{k \geq 1}$ such that $\{t_k : k \geq 1\} = S.$ Moreover, there exists a constant $C > 0$ such that for any $k \geq 1,$ we have $0 < t_{k+1} - t_k \leq C.$ (Notice that we clearly have $\lim_{k \to +\infty} t_k = +\infty$, in light of 1) above.)

3) There exists a constant $D > 0$ such that $$\limsup_{T \to +\infty} \frac{\mathrm{card}\{ k \in \mathbb{Z}_{\geq 1} : t_k \leq T\}}{T^n} \leq D.$$

Is it necessarily true that there exists a norm $\eta$ on $\mathbb{R}^n$ such that $\eta(\mathbb{Z}^n) - \{0\} = S$? (Clearly conditions 1), 2), and 3) are necessary in order for this to be true; I am wondering whether they are also, in fact, sufficient.)

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    $\begingroup$ Probably you want to require $n \geq 2$, because for $n=1$ the only allowed $S$ are arithmetic progressions $-$ but there are non-AP's that satisfy (1,2,3), such as the positive integers not congruent to $1$ or $5 \bmod 6$. $\endgroup$ Nov 6, 2019 at 1:41
  • $\begingroup$ Thank you; in fact, in my particular context, I also always assume $n \geq 2.$ $\endgroup$ Nov 6, 2019 at 1:52
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    $\begingroup$ Also condition (2) is automatic from (1) once $S \neq \emptyset$: if $S \ni t_1$ then $S$ also contains $2t_1, 3t_1, 4t_1, \ldots$, so there is never a gap longer than $t_1$. $\endgroup$ Nov 6, 2019 at 5:03

1 Answer 1


I doubt that there is a simple characterization. In any case conditions (1,2,3) are not sufficient. For example, if $n=2$ then $S$ cannot be ${\bf Z}_{\geq 100}$, and there are similar counterexamples for every $n \geq 2$, as a consequence of the following observation.

Proposition. If $\eta$ is a norm on ${\bf R}^n$, and $t_1$ is the smallest element of $\eta({\bf Z}^n)$, then for any $M>0$ there are at most $(2M+1)^n$ integer vectors $v$ with $\eta(v) \leq M t_1$.

Proof: Consider the $\eta$-balls of radius $t_1/2$ centered on all such $v$. They have disjoint interiors, and are all contained in the $\eta$-ball of radius $(2M+1)t_1/2$ about $0$. Therefore their total volume is no larger than the volume of the ball of radius $(2M+1)t_1/2$. But the volume of a ball of radius $r$ is proportional to $r^n$. Hence the total number of radius $t_1/2$ balls is at most $\left[ ((2M+1)t_1/2) \left/ \, (t_1/2) \right. \right]^n = (2M+1)^n$, Q.E.D.

Equality is attained when $\eta$ is the sup norm and $M$ is an integer. (Then we get a perfect packing of cubes.)


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