Goldbach's conjecture can't be proved to be undecidable?

Question

Conjectures concerning natural numbers which could be settled by a counterexample can, as far as I understand, not be proved to be undecidable without being proved not having a counterexample at the same time. Or?

Answer 1

If the Goldbach conjecture is undecidable (independent) of PA, then it is true for the standard model of arithmetic, i.e. the model $(\mathbb{N}, +, \times, 0, 1, <)$ with their normal interpretations.

Proof: Recall that $\mathbb{N}$ is the prime model of PA, (i.e. it embeds into every model $M$ such that $M\models$ PA). Further more, $\mathbb{N}$ is the initial segment of all models of PA. Assuming that $GC$ is independent of PA, we note that PA + $GC$ and PA + $\neg GC$ are consistent. Then, by Gödel's completeness theorem, there exists models $M_1$ and $M_2$ such that $M_1 \models$ PA + $GC$ and $M_2 \models$ PA + $\neg GC$. Suppose that $\mathbb{N} \models$ PA + $\neg GC$. Then, there exists some element $a \in \mathbb{N}$ such that $a$ cannot be written as the sum of two primes. Since $\mathbb{N}$ is the prime model of PA, we recall the fact that $\mathbb{N}$ embeds into all models of arithemtic. Therefore, if $\varphi(x)\equiv$ "$x$ cannot be written as the sum of two primes", and $\mathbb{N} \models \varphi(a)$.

By above, we showed there exist an $M_1$ such that $M_1 \models $ PA + $GC$. However, since $\mathbb{N}$ embeds into $M_1$, we note that $M_1 \models \varphi(\pi(a))$ (where $\pi$ is the embedding). However this implies that $M_1\models \neg GC$ which is a contradiction.

Thus, if $GC$ is independent of PA, then $\mathbb{N}\models GC$.