TITLE: 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?
QUESTION [10 upvotes]: 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:

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.

REPLY [7 votes]: Shortly after reading Edward H's answer, I realized that there is a self-contained compactness argument, which I missed the first time around; I thought I would present it here.
Let $T$ be an infinite set of polyominoes. Say that an infinite "polyomino" $P$ is an extension of $T$ if, for every finite region in the plane, its possible intersections with that region could also be an intersection of $t$, for infinitely many $t\in T$. (Intuitively, we're saying that local patches of $P$ look like part of a generic large element of $T$ - if you have a finite field of view, you can't distinguish it from infinitely many possibilities in $T$.)
Then, we claim that for every region $R$ which can be tiled by every element of $T$, $R$ has a tiling using extensions of $T$. (As a trivial consequence, at least one such extension must exist.)
In the original example, $T$ is the set of all $1\times n$ rectangles, and the extensions are $\{\text{infinite ray},\text{infinite line}\}$ - but since infinite lines can be tiled by infinite rays, we can just use the rays for everything.
Proof: For every positive integer $N$, let $S_N$ denote the square $[-N,N]\times[-N,N]$. Consider the restrictions of the tilings of $R$ to $R\cap S_n$ for each tile $t\in T$. Since there are finitely many ways to partition $R\cap S_n$ into tiles, but infinitely many tilings of $R$ by some $t\in T$, at least one of these partitions of $R\cap S_n$ must extend to a $t$-tiling of $R$ for infinitely many $t\in T$.
Now, fix our partition of $S_N\cap R$, and consider ways of extending it to a partition of $S_{N+1}\cap R$. Again, there are finitely many ways to do this, so again, of the infinitely many ways to extend to a $t$-tiling of $R$, infinitely many $t$ must cluster around one such partition. (The wording of this suggests a canonical choice of tiling for each $t\in T$, but we don't need to make this selection.)
Extending this construction inductively, we end up with a tiling of $R$ with tiles that have the property that every restriction of them to a finite square resembles (a part of) infinitely many tiles in $T$. Hence, all our tiles are extensions of $T$.

To show that this generality is actually good for something, here are some other special cases of the above theorem:

A region tileable by arbitrarily large squares can be tiled by infinite quadrants. (The possible extensions are quadrants, half-planes, and the whole plane, but all of these can be tiled by quadrants.)

A region tileable by arbitrarily long "zigzag" shapes can be tiled by an infinite staircase ray, i.e. the points $\{(x,y)\ |\ x,y\ge 0, x-y\in \{0,1\}\}$.

A region tileable by all "hooks" ($1\times n$ rectangles with an extra cell on the side of one end, e.g. :.......) can be tiled using an infinite ray and an infinite ray with a cell on the side of one end.


The same proof extends to most reasonable sort of grids - $\mathbb{Z}^n$, polyiamonds on the triangular grid, etc.