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I'm working through Mendelson's Introduction to Mathematical Logic and I'm having trouble proving the following statement:

'' There is no first-order theory $K$ whose models are exactly the interpretations with finite domains." (ex. 2.57)

I've been trying to argue that any theory $K$ with a model with finite domain also has a model with infinite domain, since we may extend the finite domain into an infinite one. I am however not sure how to go about this formally. I was wondering whether I am on the right track at all an if so, if anyone could help me formalise this idea.

thomjs
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  • Do you know that FOL has compactness? – ℋolo Jan 27 '19 at 21:45
  • @thomjs There are certainly theories whose models must be finite, e.g. the theory with the axiom $\forall x,y.(x=y)$. What the exercise is asking is if there is a theory for which every finite domain is a model and no infinite domains. – Derek Elkins left SE Jan 27 '19 at 21:51
  • I do, and I was actually trying to use that but couldn't quite see how. Could you give me a hint? – thomjs Jan 27 '19 at 21:51
  • It's not true that any theory with a finite model has an infinite model. In fact, you can write down an axiom in FOL with equality that says the domain has $n$ elements, so any model of that axiom will only have finite models. However, what we can't do in FOL is express the infinite disjunction "the domain has one element or the domain has two elements or..." as a collection of axioms. As Holo mentions, the standard way to show this is to use the compactness theorem. – spaceisdarkgreen Jan 27 '19 at 21:51

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Claim: "in FOL, there is no theory, $A$, such that ${\cal M}\vDash A\iff |M|<\aleph_0$"

Proof:

Let $$\varphi_n=\exists x_1\cdots\exists x_n\left(\bigwedge_{0<j<i\le n}x_i\ne x_j\right)$$ for all $n\ge2$.

In other words, a model satisfy $\varphi_n$ if it has at least $n$ elements.

Assume for the sake of contradiction that $A$ like above exists, then look at $A\cup\{\varphi_n\}_{n\in\Bbb N_{\ge2}}$, this is finitely satisfiable theory, thus it has a model, but it will have to be an infinite domain model, contradiction.

ℋolo
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