# Proof that theory of infinite numbers does not have finite model

Just curious:

Is there any proof that proves that there is no finite model of infinite numbers (Theory of infinite numbers)?

Edit: by infinite number, I mean either Reals or Naturals.

-
 What is the "theory of infinite numbers"? – Chris Eagle May 29 '12 at 23:15 @ChrisEagle I edited the question. – user1894 May 29 '12 at 23:20

Note that there is no "theory of infinite numbers", the natural numbers and the real numbers are both models of many theories. One of which is the "empty" theory in which there are no axioms, however this theory is satisfied by any mathematical structure, in particular finite structures are models of this theory.

You also need to specify the language, if we consider $\mathbb R$ only as a linearly ordered set there are things which are true and others which are false; but those are unrelated if you only consider $\mathbb R$ as a field. The language of fields does not include a symbol for $\leq$ and the language of linear orders does not include $0,1,+,\cdot$.

However we can still give this a good answer, since we can think of $\mathbb R$ as a model of RCF, the theory of Real-Closed Fields (which are provably ordered fields) and we can give the natural numbers the theory of PA, Peano Axioms.

For the real numbers note that the characteristics has to be zero (otherwise we cannot have an ordered field) and therefore: $$1\neq 0, 1+1\neq 0, 1+1+1\neq 0,\ldots,\underbrace{1+1+\ldots+1\neq0}_{n\text{ times}},\ldots$$

In a finite model this cannot happen, since we can only add $1$ to itself finitely many times before we reach some expression which we can prove is zero.

In the natural numbers we have the successor function $S$ which is described by the axioms as injective and that its image is every non-zero element. Again we can prove that there cannot be a finite model to this theory:

If there was a finite model, it would mean a finite domain for $S$, and on a finite set injectivity is equivalent to surjectivity. However $0$ is not in the image of $S$ therefore $S$ is not surjective, and in particular not injective. This cannot be true alongside with an axiom saying that $S$ is injective but not surjective.

-
 By saying "since we can only add 1 to itself finitely many times before we reach some expression which we can prove is zero," do you mean that adding 1 to itself would at some point reach zero? Or do you mean that adding being finite is a problem? – user1894 May 30 '12 at 17:32 @user1894: Consider the numbers $\{0,1,2,3,4\}$ with the usual addition and multiplication $\bmod 5$. It is not a hard exercise to see that this too is a field. However in this field $1+1+1+1+1=0$. In any ordered field we want to have $0<1$ and $a These theories satisfy the conditions $$\forall x\exists y:x>y,$$ $$\forall x:\neg(x>x),$$ and $$\forall x\forall y\forall z:(x>y\wedge y>z)\implies x>z.$$ In other words, they imply that$>$is a strict partial order without a maximal element. This cannot be in a finite set, so there is no finite model. - I am not sure what you mean by the "Theory of Infinite Numbers". In empty language (only logical symbols), you can write something like the theory of an infinite set. It has as its axioms, $$\Phi_N \equiv (\exists x_1, ..., x_n)(\bigwedge_{i,j \leq n} x_i \neq x_j)$$ That is the theory is$\{\Phi_n : n \in \omega\}$. Every model of this theory is infinite. You can also interpret the Theory of infinite numbers to mean,$Th(\omega)$of$Th(\mathbb{R})$to mean the full theory of$\mathcal{L}$structure$\omega$or$\mathbb{R}$where$\mathcal{L}$is the appropriate language, i.e. the language of arithmetics or fields. The full theory means the set the all$\mathcal{L}$-sentences that is true in$\omega$or$\mathbb{R}$respectively. Anyway, neither of these two have a finite models since$\mathbb{R}$and$\omega$models$\Phi_n$of all$n \in \omega\$. Remarkably by Upward Lowenheim-Skolem, the theory of the natural numbers and the theory of the real numbers have infinite models of arbitrarily large cardinality.

-