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(1) Some typical inclompleteness proofs use a kind of fixed point argument - for certain $\Phi$ you find a $\varphi$ with $\Phi(\mathrm{code}(\varphi))\leftrightarrow\varphi$.

(2) Some independence proofs seem to use something entirely different - you prove that given any model $\mathfrak M$ of (a theory $T$) + (some $\varphi$), you can construct from it a model $\mathfrak M_{\neg\varphi}$ of $T$ + $\neg\varphi$.

Still, these are obviously closely related: if I am not totally confused, the $\varphi$ in (2) plays the role (or even is an instance) of the $\varphi$ in (1): it provides a $\varphi$ which is non-provable (since this is the meaning of $\Phi$ in (1) and since you have a model of $\neg\varphi$ in (2)) and non-refutable (since it comes out true by a self-reference argument in (1) and since you have a model of $\varphi$ in (2)), which you seek in (1), don't you?

I thus wonder whether constructions like, say, Poincaré or Klein models of the non-Euclidean plane (a typical (2)) can be viewed as instances of application of some kind of fixed point theorems as in (1)?

Vice versa - can, say, construction of a Gödel sentence be viewed as construction, from a model, of some countermodel?

To put it still differently: in (1) we have a formula with the property that from its provability contradiction can be deduced; in (2) - a formula with the property that each of its models contains a tool to construct its countermodel. What kind of relationship exists between these two properties of formulæ?

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  • $\begingroup$ An independence proof is a bit different from what you wrote. It consist in giving a model of $T+\varphi$ and one of $T+\neg\varphi$. If you were to construct a model for $T+\neg\varphi$ out of an arbitrary model of $T$, you'd run into trouble, as you could have started with a model of $T+\varphi$ which would be troublesome. $\endgroup$
    – user1688
    Commented May 12, 2018 at 8:56
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    $\begingroup$ @Corbennick at the risk of asking a silly thing, but where lies the trouble? The newly constructed model need not be related to the original one. $\endgroup$ Commented May 12, 2018 at 10:58
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    $\begingroup$ @Andreij Bauer: Well yes, but the point is, that it is not necessary to construct a model from an existing one, that doesn't even make sense. For instance, if $T$ is the group theory and $\varphi$ is the formula $x=e$, where $e$ is the neutral element, then the only model of $T+\varphi$ is the trivial group. Then you need to construct from the trivial group a nontrivial group. How will you do that? $\endgroup$
    – user1688
    Commented May 12, 2018 at 12:39
  • $\begingroup$ @Corbennick thanks, I definitely agree that something's wrong with what I say. But could you explain more? What can I use to build models? Is it always the case that I don't need $\varphi$ itself? For example, don't I need the fifth postulate to build Poincaré model of the hyperbolic plane? Or, say, is it clear that I can perform Cohen forcing without using CH? $\endgroup$ Commented May 12, 2018 at 14:34
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    $\begingroup$ I don't agree with statement (1) in the post. Although there are some excellent famous arguments for Gödel's incompleteness theorem using fixed points, there are also other proofs that do not use fixed points. So I find it incorrect to say that "an incompleteness argument....is a kind of fixed-point argument." Model theory is full of incompleteness proofs for various theories, which proceed by showing that a theory is incomplete by exhibiting models of it satisfying different statements. $\endgroup$ Commented May 12, 2018 at 15:05

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