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I'm learning mathematical logic and I'm reading some lecture notes where the following question is asked as an exercise: Does Peano Arithmetic (PA) admit finite models?

The solution that is given in the lecture notes is the following:

No. Consider AP1 and AP2. Let M be a finite normal model and let a(0),…a(n) be the elements of the domain D, where a0 is the element of D denoted by the numeral 0. If AP1 is true in M, then any term formed by adding the symbol s (for successor) to a numeral denotes an object other than a0. This means that at most n-1 elements of D can be denoted by terms in which s occurs. But if AP2 is true in M, then for each element a(i) of D, there must be a distinct object denoted by the term formed by adding s to the numeral that denotes a(i). This entails that n elements of M are denoted by terms in which s occurs. So AP1 and AP2 can't both be true in M.

The part I don't understand is the one I put in bold: why if AP2 is true in M, then for each element a(i) of D, there must be a distinct object denoted by the term formed by adding s to the numeral that denotes a(i)?

Thank you very much for your help.

Fisher

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    I don't know if this can be translated into a proof in first-order PA, but the successor function $S: N \to N$ is injective, but not surjective. Therefore, $N$ must Dedekind-infinite. – 2017-02-03

3 Answers 3

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Any model $M$ of $AP1$ and $AP2$ will have to be infinite:

First, since $0$ is part of our language, we need to have some object in the domain $D$ of our model $M$ that $0$ denotes in $M$. Let's call this object $d_0$.

We also need an interpretation $s'$ of $s$ which maps objects from $D$ to $D$. And this must be a total function: for every object $d$ in the domain, there must be some object $s'(d) \in D$. So in particular we must have a $s'(d_0)$ object. Can this be $d_0$ itself? No, because then $AP1$ would not be true. So, $s'(d_0)$ cannot be $d_0$ itself. So, we need a second object, call it $d_1$, for which $s'(d_0) = d_1$.

OK, but now we must have some object that is $s'(d_1)$. By $AP1$ this cannot be $d_0$, but could it be $d_1$? No, that's not possible either, for then the successors of two different objects ($d_0$ and $d_1$) would be the same ($d_1$ in both cases), and that goes against $AP2$. OK, so $s'(d_1)$ needs to be a completely different object yet ... call it $d_2$.

And now you just repeat this: everytime we (have to!) introduce a new object $d_i$ and add it to our domain, we ask: buit what is $s'(d_i)$? ... and we find that it can't be any of the ones we already have, so we have to introduce a new object $d_{i+1}$, etc. etc.

So, you cannot satisfy both $AP1$ and $AP2$ using a finite model.

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If you have a finite model, for some unequal natural numbers $n,m$ you must have $S^n(0)=S^m(0)$. We might as well assume $n \lt m$. You can apply the second axiom $n-m$ times getting $0=S^{m-n}(0)$, which contradicts the first axiom.

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Hint $ $ The key idea is that the Peano axioms imply that the successor map is injective $(1$-$1)$ but not surjective (onto), so it cannot have finite models (by the Pigeonhole principle)