Let $M$ be a closed additive submonoid of $\mathbb{R}^n$ with $n\geq1$. Suppose also that there exists $r>0$ such that every ball of radius $r$ intersects $M$. I wonder if we can obtain more information on $M$. For example, if $n=1$, it is easy to see that $M$ has to be a subgroup of $\mathbb{R}$, thanks to the classical characterization of such subgroups. Is it still true when $n\geq2$?

  • $\begingroup$ "Closed" in the topological sense or is it a terminology about monoids? $\endgroup$
    – efs
    Commented Jan 16, 2020 at 13:19
  • $\begingroup$ Yes in the topological sense! $\endgroup$
    – phdstud
    Commented Jan 16, 2020 at 16:42
  • 1
    $\begingroup$ The geometric property you mentioned is known as "cobounded" (I added it in the title, you can revert if you like) $\endgroup$
    – YCor
    Commented Mar 11, 2020 at 9:23

1 Answer 1


Yes, it has to be a subgroup. Fix $v\in M$. We need to prove that $-v\in M$. It is sufficient to find an element of $M$ arbitrarily close to $-v$.

Choose $u_1,\ldots,u_n\in \mathbb{R}^n$ so that $v,u_1,\ldots,u_n$ are the vertices of a regular simplex with center at the origin. Choose large $N$ and consider the points $w_i\in M$ such that $\|Nu_i-w_i\|\leqslant r$. We may choose $N$ so large that the coordinates of $v$ in the basis $\{w_1,\ldots,w_n\}$ are negative: $v=-\sum_{i=1}^n t_i w_i,0<t_i$ (this is because $w_i/N$ is close to $u_i$ for all $i$, and the coordinates of $v$ in the basis $\{u_1,\ldots,u_n\}$ are all equal to -1.)

Now by Kronecker approximation theorem we may find a positive integer $s$ so that $st_1,\ldots,st_n$ are almost integers, denote $st_i=k_i+\varepsilon_i$, where $k_i$ are non-negative integers and $\varepsilon_i$ are small. Then $$M\ni (s-1)v+\sum k_iw_i=-v-\sum \varepsilon_i w_i$$ is close to $v$.

The closed subgroups of $\mathbb{R}^n$ are direct sums of subspaces and lattices. And the non-empty ball condition reads simply as being full-rank.


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