Tessellated space defines a recursive set? Is a space which has a regular geometric pattern necessarily a recursive set?  
It's obvious, for example, that $\mathbb{Z}^3$ is a recursive set and it has a "regular geometric pattern", so this motivated me to ask is every tessellated set recursive.   I don't have a precise definition of a ``tessellated set'' but hopefully it's somewhat clear what I mean.  (I only care about sets with one unique sort of tessellation).

It actually takes some work to make the ideas in the post precise - keep in mind that equality checking for reals is not recursive (according to the standard model of computation, anyways). So a bit of circumlocution is needed.
That said, there is a positive result here. The key is the following lemma:

Suppose $T$ is a computable subtree of $2^{<\mathbb{N}}$ (= the complete binary tree) which has a single infinite path $f$. Then $f$ is computable.

This may not seem relevant, but it applies to at least one version of your question. A bit roughly, if we have a finite set of tiles $X$ we can construe the set of partial $X$-tilings of the plane as a subtree $T_X$ of $2^{<\mathbb{N}}$. You write "I only care about sets with one unique sort of tessellation," and this corresponds to $T_X$ having a single infinite path. Consequently, in a precise sense we have:

If a finite tileset can only tile the plane in a single way, that unique tiling is recursive.

Of course I'm skipping a fair amount of detail here, but the general thrust is accurate.
