A very simple question about the relationship between incompleteness and consistency of any theory strong enough to express PA.
We know from Gödel that if such a theory is consistent, then it is incomplete (First incompleteness theorem). We also know that, if such a theory is consistent, it cannot prove its own consistency (Second incompleteness theorem).
But we also know that if a theory is inconsistent, then it is complete. So, why can't we infer from the fact that a given theory T is incomplete that T is consistent? The reasoning is that if T is inconsistent, then we could not find statements that are not provable in it.
It seems this conclusion is false because it contradicts the Second incompleteness theorem. But where's the flaw in the reasoning? Possible explanations I could come up with so far:
It is trivially the case that if a theory is inconsistent, then it is complete. But within the inconsistent theory, one could both "prove" that T is both complete and incomplete, i.e. it is false that "we could not find statements that are not provable in it". So really nothing follows from the fact a theory is provably incomplete.
The reasoning is sound; the reason this does not work is that it cannot be done in T, but is only an observation about T in a metatheory (which would again be subject to the Incompleteness theorems)?
Yes, an incomplete theory is in fact consistent.
But Gödel's theorem is a theorem about $\sf ZFC$, or $\sf PA$, and it is done in the meta-theory. And in the meta-theory, we assume that $T$ is consistent. Otherwise, there's no point in investigating it, is there?
The "fact" that $\sf ZFC$ is incomplete follows from the fact that it is consistent. Namely, the correct fact is "If $\sf ZFC$ is consistent at all, then it is incomplete".
Just a minor remark, though, there is an additional requirement for the incompleteness theorems, that the theory is recursively enumerable. Otherwise, simply take $\operatorname{Th}(\Bbb N)$, and this is a complete theory which is consistent and interprets Peano arithmetic. In fact, this is how you prove that $\operatorname{Th}(\Bbb N)$ is not recursively enumerable.