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I always thought, that when creating a theory (set of formulas of predicate logic of first order in some language) and when you want to have only infinite models, you must use infinite number of axioms. That's how Peano arithmetic or ZF are made.

But when I look at Robinson arithmetic, it is finite. Am I missing something? Does Robinson arithmetic have finite models, or I was wrong - you can "enforce" only infinite models by finite theory?

Ivan Kuckir
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    Notice that the infinite part of PA (the induction schema) isn't what makes it infinite, it is the injectivity of the successor function. – James Jun 10 '13 at 00:40

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A theory with a finite number of axioms can easily have only infinite models. One of the simplest examples is the theory of densely ordered sets. The language (over te predicate calculus with equality) has a single binary predicate symbol $R$. For the sake of familiarity we write $X\lt y$ instead of $R(x,y)$.

The axioms say that $\lt$ is a total order, which is dense. The total order axioms say that for all $x$ and $y$, exactly one of $x\lt y$, $y\lt x$, and $x=y$ holds, and that $x\lt y$ and $y\lt z$ implies $x\lt z$. In addition to the usual axioms for total order, we ask that there is more than one element, and that for any $x$ and $y$ there is a $z$ such that $x\lt z$ and $z\lt y$. This theory only has infinite models.

Or else we can use the basic theory of the successor function. Beside the usual symbols of first-order logic with equality, the language has a constant symbol $0$, and a unary function symbol $S$. The axioms say that for all $x$, $Sx\ne x$, that $0$ is not the successor of anything, and that every $x$ except $0$ has a unique predecessor. This is far simpler than Robinson Arithmetic, but only has infinite models.

There are much more complicated examples. For instance, the von Neumann- Gödel-Bernays set theory (NBG) is finitely axiomatized.

André Nicolas
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  • Thank you :) Maybe you should add, that dense order also has a "=" binary predicate. BTW is NBG in first-order predicate logic? Is "distinguishing" classes and sets unnecessary? – Ivan Kuckir Jun 10 '13 at 01:30
  • I was taking it for granted that the underlying logic is the predicate calculus with equality. As for whether NBG is first order, yes, we can define a set as a member of at least one class. So we do not need to formalize it as a two-sorted theory, though that is often done. Will change post to mention predicate calculus with equality. As you point out, one could also do it in pure predicate calculus, by adding a relation symbol and appropriate axioms. – André Nicolas Jun 10 '13 at 01:43
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Here is another example that may be of interest, especially to any who are algebraically minded.

Choose an odd prime $p$. To the first-order theory of fields (i.e., the field axioms) add

  1. $\underbrace{1+1+\cdots+1}_{p \text{ times}}=0$
  2. $\forall x \exists y (x = y^2)$

Now you have the first-order theory of characteristic $p$ fields that are quadratically closed. The algebraically closed field of characteristic $p$ is an infinite model, but there can be no finite models, since the squaring map is not injective, and hence not surjective.

Louis Deaett
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