I understand that every finite set is recursively enumerable, as I see that one could just encode each element of some finite set on a Turing Machines tape, and then have the machine check each member against any input to determine set membership....


...However, it isn't clear to me how there is an analog to this method in the domain of Diophantine representation, even though Matiyasevich's theorem assures us that one does exist. (That is, a set is recursively enumerable iff it is Diophantine. Every finite set is recursively enumerable, thus also Diophantine.) 


In your answer, please give the explicit method by which any finite set can be written in a Diophantine representation.



Why is every finite set Diophantine?
Or, very simply stated, given the finite set $S = \{a\_1, \dots , a\_k\}$, consider the diophantine equation: $$(n-a\_1)\dots(n-a\_k)=0.$$
EDIT: Then we can write S as $\{ \ n \ | \ \exists x : (n-a\_1)\dots(n-a\_k)=0\ \}$. (Thanks David)


