CH for tilings of the plane Given any set of jordan curves that can tile the plane, how to prove that the number of possible tilings using tiles from this set is either in bijection with the real numbers or a (possibly infinite) subset of the integers?
Two tilings are equal if they can be made to coincide by translations or rotations.
 A: Are your tiles square shaped? One can then prove the result by what is essentially a compactness argument. Here is a brief idea:
Tile in order a square of size $1\times1$, then a larger square containing that one, of size $2\times 2$, then a larger one containing it, of size $3\times 3$, etc. 
Suppose that your tiling allows us you to tile the plane in a non-periodic fashion. Then, for some $n$, you will have at least two options on how to tile the $n\times n$ square when you get there. Continue "on separate boards" with each of these two ways. Again, by non-periodicity, you should in each case reach a larger $m$ such that the $m\times m$ square can be tiled in at least two ways when you get there (of course, the $m$ in one case may be different from the $m$ in the other case). Continuing "on separate boards" in this fashion, you are building a complete binary tree, each path through which gives you a "different" tiling of the plane. The quotes are here, as we are not yet distinguishing between translations. But there are only countably many translates of a given tiling (if we insist that, say, the borders of our squares coincide with the lines of the form $x=n$ or $y=m$ for $n,m\in{\mathbb Z}$). But then, even after identifying translates, we are left with continuum many different tilings of the plane.
The other possibility is that your tiles does not allow non-periodic tilings. Then, no matter what path you follow, the process above should stop "splitting". But then you are only producing countably many tilings in this fashion.
