$2^{\aleph_0}$ complete consistent extensions of the arithmetic

Question

I'm asked to prove, using Gödel's incompleteness theorem (version every consistent and recursively axiomatizable extension of arithmetic is incomplete) that there are $2^{\aleph_0}$ complete consistent extensions of the arithmetic. I know how to prove this using ultraproducts and the powerset of prime numbers but I cannot get the proof that uses Gödel's theorem.

Answer 1

A proof that uses Gödel's theorem is the following :

Let $PA$ be the usual theory of arithmetic. By Gödel's incompleteness, any consistent finite extension of it is incomplete (this is true for larger classes of extensions, but here finite will suffice).

Let $\{\varphi_n\mid n\in \mathbb{N}\}$ be an enumeration of the set of all sentences of arithmetic.

We construct $T : 2^{<\omega} \to \{S\mid S$ is a consistent theory that finitely extends $PA\}$ by recursion : $T_\emptyset = PA$, $T_{s\frown 0} = T_s \cup \{\varphi_n\}$ where $n$ is the least integer such that $\varphi_n$ is independent of $T_s$ (such $n$ exist because $T_s$ is a finite extension of $PA$ and is thus incomplete). $T_{s\frown 1} = T_s \cup\{\neg \varphi_n\}$ with the same $n$.

By construction, for each $s\in 2^{<\omega}$, $T_s$ is a consistent extension of $PA$.

Now if we let $T : 2^\omega \to \{S\mid S$ is a consistent theory that extends $PA\}$ be defined by $T(s) = \displaystyle\bigcup_{n<\omega } T_{s\mid n}$ we clearly get an injective map. This proves the claim