TITLE: Sudoku on a countably infinite board
QUESTION [26 upvotes]: Consider the game of Sudoku played on an infinite board where the subsquares are also infinite, i.e. our board is indexed by $\mathbb{N}^2 \times \mathbb{N}^2$. Let's call a solution to such a game a function $f(a, b, m, n)$ which assigns a natural number to each space $(m,n)$ in each subsquare $(a,b)$, such that each row, column, and subsquare contains every natural number exactly once.
It is clear that such a solution exists, as for any finite board state, given any natural number and any row, column, or subsquare, there are always at most a finite number of "collision" squares, and so with infinite spaces at our disposal we can always pick a space to put this number in, and continue to do this infinitely until we have filled the board.
However, I'm having trouble constructing an explicit example of such a solution, which does not rely on this choice-like magic. My initial thought was to use products of primes  to guarantee that you don't have a collision, but while I can get plenty of solutions with no repetitions, guaranteeing that every row, column, and subsquare contains every label seems like a lot harder of a challenge. But, I suspect I'm missing  a very elegant / basic solution. Any ideas / hints?

REPLY [6 votes]: Just to make clear, you do not have to be 'clever enough' to come up with a 'closed-form' solution. Having 'finitely many obstructions' is enough to rigorously prove the existence of a solution. Here is one way.
The board is divided into big-rows, each of which has infinitely many small-rows. Same for big-columns and small-columns. Call the $j$-th small-row in the $i$-th big-row row $(i,j)$, and similarly for column $(i,j)$. The board is also divided into subsquares, each of which is the intersection of a big-row and a big-column. Call the intersection of the $i$-th big-row and $j$-th big-column square $(i,j)$.
There are a few requirements we need to satisfy besides no collisions:

$R(i,j,k)$: Row $(i,j)$ has an occurrence of $k$.
$C(i,j,k)$: Column $(i,j)$ has an occurrence of $k$.
$S(i,j,k)$: Square $(i,j)$ has an occurrence of $k$.
$D(i,j,k,l)$: The cell $(i,j,k,l)$ is filled.

In each round $m∈ℕ$, we can satisfy all the requirements $R(i,j,k)$, $C(i,j,k)$, $S(i,j,k)$, $D(i,j,k,l)$ where $\max(i,j,k,l) = m$ by handling them one by one, since there are only finitely many of them. To see why we can succeed for each requirement, note that at any point we have filled only finitely many cells. Thus for each of the $R,C,S$ requirements we can easily pick an empty cell to satisfy it. And for each $D$ requirement we can easily pick a natural number that we have not used yet. We can even make this choice deterministically, by picking for each of the $R,C,S$ requirements the uppermost then leftmost cell that works, and by picking for each $D$ requirement the smallest natural number that works.
At the end, we have satisfied all the requirements and still have no collisions, and hence have a filled infinite sudoku board as desired.