Since arithmetic has a model (thus it is consistent) why care if consistency can't be proved?

Question

Since arithmetic has a model, the numbers as we know them, it is consistent. Why do we care if consistency can't be proved within arithmetic? Do I miss something, ie in what we can consider a model?

Answer 1

This is a good question. The answer comes in two parts.

Mathematically speaking it is more important that we understand that $\sf PA$ is incomplete. So there will always be true statements that we cannot prove just from $\sf PA$. And that whatever reasonable theory (read: satisfies the condition of the second incompleteness theorem) we use as a foundational theory is also incomplete.

This is an important understanding, and the statement "$T$ is consistent" for whatever suitable $T$ we use is just an excellent example of this incompleteness phenomenon.

More specifically, the statements about the consistency of theories allow us to establish an order between theories. One theory is stronger than another if it proves its consistency.

It is true that $\sf PA$ is consistent if we assume that $\sf ZFC$ is consistent. But what about $\sf ZFC$ now? So we work in $\sf ZFC$ extended by additional axioms, then what about those axioms? are they consistent? So we add more and more and more, and this becomes a cat and mouse game.

The easy way out is to say that we are simply "manipulating strings", but this is just a way of saying that we work within a theory which is some subtheory of $\sf PA$ that can internalize logic. So why is that theory consistent again?

So we have some sort of infinite descent here, which is not ideal. The second incompleteness theorem tells us that this descent is in fact infinite. Whatever theory you have, if it is good enough for a computer to have a proof verifier algorithm for the theory, and you can do "enough" arithmetic in it, then it cannot prove its own consistency.