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I am trying to show that compactness doesn't apply to infinite languages like $L_{\omega_1,\omega}$ that allow infinite FOL sentences like $\forall x\, \bigvee_{n\in \omega} x \approx S^n(0)$, i.e. $\forall x\, (x \approx 0 \vee x \approx S(0) \vee x\approx S(S(0)) \dots)$.

I already picked a language $L=(0,S)$, gave an example of a set of sentences which is satisfiable, since every subset $\Gamma_m$ of $\Gamma_n\colon x\approx 0 \vee \dots \vee s\approx S^n(0)$, has a finite model for every $m$ smaller than $n$.

I am trying to show that nonetheless $\Gamma_n$ has no model. I can't get my head around how to perform this last step. Maybe my example of a set of sentences $\Gamma_n$ which is supposed to be finitely satisfiable but not satisfiable is not the right one?

Alex Kruckman
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JanikG.
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  • What does $W$ represent? I guess $0$ is a constant symbol and $S$ is a function? Please explain all your notations (in general). – A. Pongrácz Jan 08 '19 at 19:49
  • Sorry, I am new to formatting. W with a subscript n is supposed to represent the infinite disjunction. – JanikG. Jan 08 '19 at 19:52
  • And yes, 0 is a constant symbol and S is a function. Sorry again for ugly formatting. – JanikG. Jan 08 '19 at 19:53
  • I edited your post to include MathJax formatting. You should use this formatting in future posts on this site. Here's a tutorial / reference on how to do this: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Alex Kruckman Jan 09 '19 at 15:21
  • "Every subset $\Gamma_m$ of $\Gamma_n$... has a finite model for every $m$ smaller than $n$" doesn't make any sense to me. It would make more sense notationally to let $\Gamma = {\Gamma_n\mid n\in \omega}$, and show that $\Gamma$ is finitely satisfiable by arguing that for every $n$, the set ${\Gamma_m\mid m<n}$ has a model. (Whether this model is finite is irrelevant.) But in any case, your choice of $\Gamma_n$ doesn't work. – Alex Kruckman Jan 09 '19 at 15:24

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Actually, you don't need any signature: $L=\{=\}$ is completely fine. Note that you can express the property for any $n\in \mathbb{N}$ that the model contains at least $n$ different elements. In fact, there is a finite first-order formula that expresses this. Let $\phi_n$ be such a formula.

(E.g., $\phi_3$ is $\exists x_1,x_2,x_3 \,\, x_1\neq x_2 \wedge x_2\neq x_3 \wedge x_1\neq x_3$.)

The theory $T$ now consists of

$\bullet$ all the $\phi_n$,

$\bullet$ and the formula that says that one of the $\phi_n$ must be false (this is an infinite disjunction of all the negations of the $\phi_n$)

Altogether, this means that the model is infinite and finite (contradiction).

But any finite subsystem is satisfiable (by some finite set).

A. Pongrácz
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