My question probably needs to include the definitions of the terms in the title so I will first ask the question and then introduce the necessary definitions.

The following Theorem is stated without proof here.

Let $T$ be a tiling of the space $\mathbb{R}^n$. If $T$ is a repetitive tiling, then $T$ has finite local complexity.

However, I think that there is a periodic (and so repetitive) tiling of the line $\mathbb{R}$ which does not have finitely many tile types, and so does not have finite local complexity, contradicting the statement. The example is given by the tiling $$T=\{t_{n,m}=[n+(1/2)^{m+1},n+(1/2)^{m}]\:|\:n\in\mathbb{Z},m=0,1\ldots\}.$$ $T$ is obviously a tiling of $\mathbb{R}$ as the union of all the $t_{n,m}$ cover the line, and the intersection of the interior of any two tiles is empty, and $T$ clearly has period 1 as $t_{n,m}+1=t_{n+1,m}$ and so is repetitive. Finally, $T$ has infinitely many tile types given by varying the value of $m$ for fixed $n$ and so $T$ does not have finite local complexity.

Have I missed something obvious in the definitions below, or my analysis of the above example, or is this really a counterexample to the stated Theorem? I think a clear fix would be if one restricted the definition of a tiling to those tilings, $T$ of $\mathbb{R}^n$ such that the intersection of any compact subset of $\mathbb{R}^n$ with the tiles $t$ in $T$ is non-trivial for only finitely many $t$. In this case, it seems clear that the Theorem is true.

The necessary definitions are below.

Tiling link

We say the set $t\subset\mathbb{R}^n$ is a tile if $t$ is a non-empty compact subset of $\mathbb{R}^n$ such that the closure of the interior of $t$ is equal to $t$. We say a set of tiles $T=\{t_{\lambda}|\lambda\in\Lambda\}$ is a tiling of $\mathbb{R}^n$ if

  • $\bigcup_{\Lambda} t_{\lambda}=\mathbb{R}^n$, and
  • for all $\lambda\neq \lambda'$, we have $\mbox{Int}(t_{\lambda})\cap \mbox{Int}(t_{\lambda'})=\emptyset$.

Repetitive link

Let $T$ be a tiling of $\mathbb{R}^n$. We say $T$ is repetitive if, for all $r>0$, $x\in\mathbb{R}^n$, there exists a $\delta\in\mathbb{R}$ such that, for any $y\in\mathbb{R}^n$, there is some $z\in\mathbb{R}^n$ with $B_{r}[T,x]\sim B_{r}[T,z]$ and $|z-y|<\delta$.

Here, $B_{r}[T,x]=\{t\in T\:|\:t\cap B_r(x)\neq\emptyset\}$ is the set of tiles in $T$ which have non-trivial intersection with the open ball $B_r(x)$ of radius $r$ and centre $x\in\mathbb{R}^n$. We say two sets of tiles are $\sim$-equivalent if a translation of $\mathbb{R}^n$ exists which takes one set to another.

Finite local complexity link

We say that $T$ has finite local complexity if, for all $r>0$ the set $\{B_r[T,x]\:|\:x\in\mathbb{R}^n\}/{\sim}$ is finite.

Note that if $T$ does not have finitely many tile types, then $T$ does not have FLC. By tile types, we mean the $\sim$-equivalence classes of single tiles in $T$.

  • $\begingroup$ With these definitions, there are no repetitive tilings: Let $x\in\mathbb R^n$ and $r>0$ be arbitrary, let $t$ be a tile with $x\in t$. As $t$ is compact, it has finite diameter and $t\cap B_r(z)=\emptyset$ if $|z-x|>\operatorname{diam}(t)+r$. No matter what $\delta>0$ we choose, we can find $y$ with $|y-x|>\operatorname{diam}(t)+r+\delta$ and then $t\in B_r[T,x]$ and $t\notin B_r[T,z]$ for all $z$ with $|z-y|<\delta$. Your definition does not even resemble what you link to. $\endgroup$ May 2, 2013 at 15:28
  • $\begingroup$ I think that you mean $B_r[T,x]\sim B_r[T,z]$ in your definition of repetitive. $\endgroup$ May 2, 2013 at 15:30
  • $\begingroup$ I do, thanks for the spot. $\endgroup$
    – Dan Rust
    May 2, 2013 at 15:31
  • $\begingroup$ Indeed, it looks like one needs some extra condition, such as local finteness (i.e. each point has an open neighbourhood that intersects only finitely many tiles) $\endgroup$ May 2, 2013 at 15:40

1 Answer 1


I'm the author of the page you refer to. Yes, you are right: One needs to require locally finite. I will change the text on the page you mention. In carefully written papers or books this theorem is stated as follows: "Let $\Lambda \subset \mathbb R^d$ be a locally finite point set that is repetitive. Then, $\Lambda$ is also FLC." (see Baake, Grimm: "Aperiodic Order" Cambridge University Press 2013, Proposition 5.5, p. 140, where you find also a proof)

  • $\begingroup$ Many thanks for the response and reference. On a personal note, I would also like to thank you for contributing to the wonderful resource that is the tilings encyclopedia. It's been invaluable in my studies so far. $\endgroup$
    – Dan Rust
    Nov 18, 2013 at 14:12

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