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Am I right in thinking that there are countably infinitely many nonequivalent unprovable statements in ZFC due to the Gödel's first incompleteness theorem? If not, why?

user132181
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    What do you mean by "equivalent" ? You can consider the sequence of theories $T_1=ZFC, T_{n+1}=T_n+(T_n$ is consistent $)$. Then each term in the sequence is strictly stronger than the preceding one. – Ewan Delanoy Oct 23 '15 at 06:30
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    If there were only finitely many undecidable statements in ZFC, wouldn't it follow that some finite extension of ZFC is complete? – bof Oct 23 '15 at 06:38

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First of all, "countably" is redundant, because there are only countably many sentences in the first place. With that out of the way, you are correct. Assume there are only finitely many unprovable statements (up to equivalence modulo ZFC) $\sigma_1 \ldots \sigma_n$, and that ZFC is consistent. Then take $\text{ZFC}^\star$ some complete, consistent extension of ZFC. Since there are only finitely many $\sigma_i$, we can effectively axiomatize $\text{ZFC}^\star$ by taking the axioms of ZFC together with one of $\sigma_i$ or $\neg \sigma_i$ for each $i$ (depending on whether $\text{ZFC}^\star \models \sigma_i$), contradicting the first incompleteness theorem.

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    In fact, even more is true: there is an infinite family of sentences $\sigma_i$ such that, for any $X\subseteq\mathbb{N}$, the theory $ZFC\cup{\sigma_i: i\in X}\cup{\sigma_j: j\not\in X}$ is consistent - but this takes more work. – Noah Schweber Oct 23 '15 at 07:31
  • @NoahSchweber: I think you've missed out a negation or something: your theory as described doesn't depend on $X$. – Rob Arthan Oct 23 '15 at 22:43
  • @RobArthan of course you're right, the third term is supposed to be "${\neg\sigma_j: j\not\in X}$." – Noah Schweber Oct 24 '15 at 00:28