Tiling arbitrarily large portions of the plane implies tiling the plane?

Question

Suppose that one has a finite collection of types of (polygonal) tiles. We say that we can cover a region of the plane if we can place tiles in a non-overlapping fashion (except for edges of tiles touching) such that the region is contained in the union of the tiles.

Q: Suppose that, with our tiles, we can cover any compact subset of the plane. Does it follow that we can tile the entire plane?

The statement does not hold if we have an infinite set of tile types. For example, we could take arbitrarily large squares that are missing small notches on the side. Any region can be covered with a large enough square, but we have no way to fill in the missing notch.

For context, I'm trying to understand non-periodic tilings, and I'm not sure how one could generically show that a given collection of tiles non-periodically tiles the plane unless a criterion similar to this one is true.

Answer 1

A compactness argument should do the trick. I assume each tile type has positive area.

A configuration of tiles is specified by giving a type, position (of a given reference point) and angle of rotation for each tile. Since there are only a finite number of tiles and each has positive area, only finitely many tiles can intersect a bounded region. Allowed configurations of tiles with reference points in, let's say, a closed unit square form a compact metric space $K$. Tiling the plane with such unit squares, we get a configuration space $\Omega$ for the infinite plane that is a closed subset of the cartesian product of copies of $K$. By Tychonoff's theorem this is a compact metrizable space. Given sequence of configurations $\omega_n$, each consisting of a finite arrangement of non-overlapping tiles that cover, say, $[-n,n] \times [-n,n]$, some subsequence will have a limit $\omega$, and that limit must be a tiling of the whole plane.

Answer 2

Robert Israel gives a nice answer using Tychonoff's theorem.

Here is an argument which uses diagonalization directly, avoiding Tychonoff's theorem.

Let the set of tiles be denoted $t_1,\ldots,t_K$. Let $\Delta$ be an upper bound to $diameter(t_k)$. By translation we may think of each $t_k$ as specific subset (polygon) in $\mathbb{R}^2$ containing the origin; let me refer to these specific tiles as the "prototiles". So in a general tiling, each of the tiles is a copy of some prototile $t_1,\ldots,t_K$ obtained by translation and rotation.

For each natural number $n$ let $\mathbb{T}(n)$ denote a finite tiling that covers $B(n+\Delta)$ (the closed ball centered on the origin of radius $n+\Delta$). Some tile in $\mathbb{T}_n$ contains the origin. So after translating $\mathbb{T_n}$ a distance at most $\Delta$ and then rotating $\mathbb{T_n}$ around the origin we may assume that $\mathbb{T_n}$ contains one of the protitles $t_1,\ldots,t_K$.

After these translations and rotations, although $\mathbb{T}(n)$ need no longer cover $B(n+\Delta)$, it still covers $B(n)$. In fact, $\mathbb{T}(n)$ has a finite subtiling that covers $B(n)$ and is contained in $B(n+\Delta)$, namely the union of all tiles in $\mathbb{T}(n)$ that intersect $B(n)$.

There are only finitely many of $t_1,\ldots,t_K$. So, there is a subsequence $n^0_1 < n^0_2 < n^0_3 < \cdots$ such that each of $\mathbb{T}(n^0_i)$ contain the same prototile.

In general the following statement is true for each natural number $n$:

$(*)_n$ There are only a finite number of tilings which contain a given prototile, which cover $B(n)$, and which are contained in $B(n+\Delta)$.

Using $(*)_1$, the sequence $(n^0_i)$ has a further subsequence $n^1_1 < n^1_2 < n^1_3 < \cdots$ such that each of $\mathbb{T}(n^1_i)$ has a common subtiling that covers $B(1)$ and is contained in $B(1+\Delta)$.

Using $(*)_2$, the sequence $(n^1_i)$ has a further subsequence $n^2_1 < n^2_2 < n^2_3 < \cdots$ such that each of $\mathbb{T}(n^2_i)$ has a common subtiling that covers $B(2)$ and is contained in $B(2+\Delta)$.

$\vdots$

By induction, we obtain an infinite nested sequence of subsequences $n^j_1 < n^j_2 < n^j_3 < \cdots$ such that for each $j$, each of the tilings $\mathbb{T}(n^j_i)$ has a common subtitling that covers $B(j)$ and is contained in $B(j+\Delta)$.

By diagonalization, we may extract a sequence $n(1) < n(2) < n(3) < \cdots$ which is simultaneously a subsequence of each of $n^j_1 < n^j_2 < n^j_3 < \cdots$. It follows that for each $m \ge 1$ the tilings $\mathbb{T}(n(m))$ all contain a common subtiling $\mathbb{T}_1$ that covers $B(1)$, for $m \ge 2$ they all contain a common subtiling $\mathbb{T}_2$ that covers $B(2)$, for $m \ge 3$ they all contain a common subtiling $\mathbb{T}_3$ that covers $B(3)$, ...

The union of the nested subtilings $\mathbb{T}_1 \subset \mathbb{T}_2 \subset \mathbb{T}_3 \subset \cdots$ is the desired tiling of the whole plane.