# Math Genius: Definition of structure and satisfaction on Set theory(ZFC).

I’m studying mathematical logic with Enderton’s book. Afther reading the definition of structures, satisfaction and models, I have thought that the definition really take set theorical notions such as functions. Is this only one way to define structures and satisfaction? Can we define these model theorical notions without set theory?

Thanks.

This is a somewhat vague question, but there are various precise results which indicate a negative answer. However, not a lot of set theory is needed here.

The most obvious one is Tarski’s undefinability theorem. Roughly speaking, there are lots of very nice functions assigning natural numbers to sentences in the language of arithmetic (Godel numberings); Tarski says that for any such map $$ulcornercdoturcorner$$ the set $${ulcornervarphiurcorner: mathbb{N}modelsvarphi}$$ is not a definable subset of $$mathbb{N}$$. (Here “$$mathbb{N}$$” refers to the structure $$(mathbb{N};+,cdot)$$.)

So if I want to talk about the satisfaction of a first-order sentence in $$mathbb{N}$$, the first-order logic of $$mathbb{N}$$ itself isn’t sufficient. Moreover, Tarski is very robust: we can replace $$mathbb{N}$$ with lots of other structures, which also admit very nice Godel numberings, and get the same result.

A more technical example concerns absoluteness. We can implement model theory inside a sufficiently rich theory, like $$mathsf{ZFC}$$. In this context we can ask how “stable” assertions of satisfaction are. In particular:

• Suppose $$M$$ is a model of $$mathsf{ZFC}$$, $$A$$ is a structure in $$M$$, and $$varphi$$ is a first-order sentence such that $$M$$ thinks $$Amodelsvarphi$$. Now suppose $$N$$ is an “extension” of $$M$$; does $$N$$ still think $$Amodelsvarphi$$?

The answer to this question is yes, but less robustly than one might hope. Consider the following variant:

• Suppose $$M$$ is a model of $$mathsf{ZFC}$$, $$A$$ is a structure in $$M$$, and $$varphi$$ is something $$M$$ thinks is a first-order sentence such that $$M$$ thinks $$Amodelsvarphi$$. Now suppose $$N$$ is an “extension” of $$M$$; does $$N$$ still think $$Amodelsvarphi$$?

The point is that if $$M$$ is a non-$$omega$$-model of $$mathsf{ZFC}$$, then there will be things in $$M$$ which $$M$$ thinks are first-order sentences but which actually aren’t (intuitively, they’re secretly infinitely long). This added level of flexibility changes the situation dramatically: a negative answer to the broader question was established by Hamkins and Yang.

To see how this is relevant, we need to unpack the term “extension.” Basically, we’re looking at extensions of the given model of $$mathsf{ZFC}$$ which don’t impact the elements of $$A$$ itself: e.g. they don’t add new elements to $$A$$, or change how the existing elements of $$A$$ interact with each other. So all the changes which occur when we pass from $$M$$ to $$N$$ are “higher-order” as far as $$A$$ is concerned: the collection of functions over $$A$$ changes, but $$A$$ itself doesn’t change.

That said, only a mild amount of set theory is needed to talk about satisfaction of a given sentence in a given structure. For example, this can be measured by Turing degrees: if $$mathcal{X}$$ is a structure with domain $$mathbb{N}$$, the theory of $$mathcal{X}$$ has Turing degree at worst $$(AtDiag(mathcal{X}))^{(omega)}$$, the $$omega$$th jump of the atomic diagram of $$mathcal{X}$$.

This can also be thought of from the point of view of the axiomatic strength required to prove basic things about $$models$$; at least for countable structures, this is ably handled by reverse mathematics, and the relevant theories (variously $$mathsf{RCA_0}, mathsf{WKL_0},mathsf{ACA_0}$$, and $$mathsf{ACA_0^+}$$; see the discussion here) are fairly weak.

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