Question about Finite Model on Robinson Arithmetic

Question

So I was supposed to create a finite model for Robinson Arithmetic in an exam and show that it was a finite model, but I was unable to do so. Would appreciate any help with this problem because I feel endlessly frustrated at my attempt at this problem! Looking at my notes again, this shouldn't be possible.

Answer 1

I forgot to answer this question yesterday but I am answering it here for sake of completion. In particular, there do not exist any finite models for Robinson Arithmetic (or the Peano axioms or anything similar) due to the axioms that 0 cannot be a successor of any number and that the successor function is injective. Notice that if there were a finite model for Robinson arithmetic, then, since injectivity implies surjectivity on a finite set of same cardinality, it implies a finite domain for the successor function. But this is impossible because $0$ is not in the image of S, so it can't be surjective and thus injective, a contradiction.

Another reason is that on any finite model, it would not be possible to show that the successor of the last element is not $0$.