[Math] If a subset of the square grid can be tiled by $1\times n$ rectangles for every $n$, can it be tiled by infinite rays

geometrylogicpolyominotiling

Suppose that we have some set $S$ of grid-aligned squares in the plane; equivalently, we can think of our set as $S\subset \mathbb{Z}^2$. Suppose that for every positive integer $n$, $S$ can be tiled by disjoint copies of a $1\times n$ rectangle (possibly rotated). Is it necessarily the case that $S$ can be tiled by congruent copies of an infinite "ray" of squares, i.e., a region congruent to the set $\{(k,0)\ |\ k\ge0\}$?

For example, the region given by removing from $\mathbb{Z}^2$ the points $(0,0), (1,1)$, and $(-1,2)$ can be tiled by every $1\times n$ rectangle, and also by infinite rays, as shown:

enter image description here

I'd like to say that this follows from some kind of compactness argument, but my attempts to apply such reasoning ran into trouble (I can't rule out that the $1\times n$ tilings have their endpoints move away from the origin without a well-defined limiting behavior). Any thoughts on how to prove this (or a counterexample, if my intuition is badly wrong)?

I believe the same statement should hold in $\mathbb{Z}^n$ with $1\times1\times\ldots\times1\times n$ tiles and the same kind of rays; if the original statement proves easy to show, I would be interested in this generalization.

As a remark, the converse is obviously true, since the infinite ray can be tiled by any $1\times n$ rectangle.

Best Answer

Apologies in advance for shooting sparrows with cannons, but I can't give away such a funny opportunity to see "pure" math being applied to "simple" geometry problems.

I'll be using Gödel's compactness theorem, which is well-explained here, so check it out. In essence, the theorem is as follows: In first-order logic, if a property is witnessed by arbitrarily large finite numbers, then there is a way to witness it by infinities. It certainly feels relevant in the post's context.

Consider the theory of the abelian group $\mathbb{Z}^2$ with generators $a,b$. Thus the following sentences hold:

  • $ab=ba$
  • $\forall y,\,ey=ye=y$
  • $a\ne b\ne e$
  • $a^{-1}\ne a^{1729}$
  • etc.

We collect all valid sentences into the set $\operatorname{Th}(\mathbb{Z}^2,a,b)$.

We introduce the relation $R(x,y)$, which says that "the group elements $x,y$ of $\mathbb{Z}^2$ belong to the same tile". Then the following things can be expressed as formulas/sentences:

  • $R$ is an equivalence relation (left as exercise)
  • Every tile is a single strip of rectangle, or equivalently, any two elements not on a same row/column cannot be in the same tile. We express this with infinitely many sentences. For example, $\forall x,\lnot R(x,a^{-3} b^4 x)$ would be one of them. We also need the rectangles to be connected, and they can also be expressed by infinitely many sentences, for example, $\forall x,R(x,a^6 x)\rightarrow \bigwedge_{i=1}^5 R(x,a^i x)$.
  • For a particular group element $x$, we can ask for it not to be tiled. We use the sentence $\phi(x)$, which says that $\forall y\ne x,\lnot R(x,y)$.
  • For a particular group element $x$, we can also ask for it to be tiled, and we get to choose how big the tile is! For an integer $n>0$, we use the sentence $\psi_n(x)$ to mean that $x$ is in a tile of size $\ge n$, which is expressed as: $$\left(\bigvee_{i=1-n}^0\bigwedge_{j=i}^{i+n-1} R(x,a^j x)\right)\lor\left(\bigvee_{i=1-n}^0\bigwedge_{j=i}^{i+n-1} R(x,b^j x)\right)$$

Now consider making a theory $T$ in the language of groups with the marked elements $a,b$ and the relation $R$. The consistency of this theory will claim that $S\subseteq \mathbb{Z}^2$ can be tiled by infinite-sized tiles. It will contain all appropirate axioms listed above, but we take special care when putting in any $\phi(x)$ or $\psi_n(x)$ into $T$. In particular, if $x\in S$ is an element of $\mathbb{Z}^2$ expressed as a word in $a,b$, then we put $\psi_n(x)\in T$ for all $n$ (but of course $\phi(x)\notin T$). Otherwise if $x\notin S$, then we only put $\phi(x)\in T$.

By the compactness theorem, the consistency of $T$ will follow from the consistency of every finite segment $T_0\subseteq T$. However as $T_0$ is finite, there is some bound $N\in\mathbb{N}$ such that every $\psi_n(x)\in T_0$ will be of the form $n\le N$. In particular, $T_0$ is now a segment of a theory whose consistency claims that $S$ can be tiled by rectangular tiles, and every tile has size at least $N$. By our assumption, we can simply tile $S$ by rectangular tiles of size $N$, and see that $T_0$ is indeed consistent. By the compactness theorem, $T$ is consistent.

Thus we know that $S$ can indeed be tiled by rectangles of infinite size. Any such rectangle is either a ray or can be tiled by two rays. Hence $S$ can also be tiled by rays.

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