TITLE: Tiling $\mathbb{R}^2$ with polyomino
QUESTION [6 upvotes]: Divide $\mathbb{R}^2$ into disjoint unit squares, let Q be a tile (a polyomino, a finite set of connected unit squares). If, for every finite set S of unit squares in $\mathbb{R}^2$, - I can find a finite set of disjoint Q-tiles (a tiling) such that S is a subset of this tiling, how to prove that I can tile entire $\mathbb{R}^2$ with Q-tiles, (without axiom of choice)?

REPLY [11 votes]: I know three different proofs of this.

First, you can appeal to König's lemma. View the set
of tilings of increasingly large squares based at the
origin (that is, they minimally cover such a square) as a
tree under the subtiling relation. This is a finitely
branching tree, since any tiling of a square has only
finitely many extensions to cover the next larger square.
But your assumption says that this tree is infinite. Thus,
it has an infinite branch, by König's lemma. Such a
branch gives a tiling of the entire plane.
Second, you can view the previous argument as a
compactness argument, if you consider the right topology.
Third, one can argue from nonstandard analysis. Take a
nonstandard model of the natural numbers. By the transfer
principle, there must be a tiling of some nonstandard
size square using tiles from your tile set. But a nonstandard
(pseudo)finite square includes many actual copies of the
standard plane inside it, around the nonstandard numbers in the
very center of the nonstandard square. Thus, there is a
tiling of the actual standard plane.