Must heredity be explicilty stated in addtion to Peano's axioms when defining natural numbers?

Question

My question is stated in bold text.

This question pertains to the following formulation of Peano's axioms used to formalize an introduction of the natural numbers (beginning with 1) consisting of successions of vertical strokes drawn on paper in horizontal rows.

• I. $1$ is a number.
• II. To every number $a$ there corresponds a unique number $a^{\prime},$ called its successor.
• III. If $a^{\prime}=b^{\prime},$ then $a=b.$

• IV. $a^{\prime}\ne1$ for every number $a.$

• V. Let $A\left(x\right)$ be a proposition containing the variable $x.$ If $A\left(1\right)$ holds and if $A\left(n^{\prime}\right)$ follows from $A\left(n\right)$ for every number $n,$ then $A\left(x\right)$ holds for every number $x.$

Are these axioms sufficient to determine the natural numbers $\mathbb{N}\equiv\left\{ 1,2,3,\dots\right\} ?$ Or is it necessary to add a further requirement that a number must inherit its membership in $\mathbb{N}$ from $1?$

For example, it seems that the set $\mathcal{N}=\mathbb{N}\cup\left\{ h\vert i\in\mathbb{Z}\land h=i+\frac{1}{2}\right\} $ is consistent with these axioms, with $a^{\prime}\equiv a+1,$ in the standard sense.

I propose that the following statement would fulfill such a requirement: The set of natural numbers contains the number $1$ as well as every number reached successively beginning with $1,$ with succession conforming to Peano's axioms as stated.

That avoids explicit reference to a "successor function", which would require additional definitions. Nonetheless, any such requirement stated in addition to Peano's axioms seems to presuppose ordering. Terms such as beginning with, repeated application, successively, inherited from, etc., appeal to our a priori notions of ordering. So I'm left wondering to what extent our subsequent theorems pertaining to ordering are actually derived results.

Answer 1

Yes, the (first-order) theory PA is not sufficient for characterizing $\mathbb{N}$ up to isomorphism.

In fact, no first-order theory can do the job for any infinite structure at all! This is a consequence of the compactness theorem:

• Suppose $\mathcal{A}$ is an infinite structure in a language $\Sigma$ and $\Gamma$ is a set of sentences true in $\mathcal{A}$.

• Fix a cardinal $\kappa>\vert\Gamma\vert$, and let $\Sigma^*$ be the language gotten by adding $\kappa$-new constant symbols $\{c_\eta:\eta<\kappa\}$ to $\Sigma$.

• Let $\Gamma^*=\Gamma\cup\{c_\eta\not=c_\theta:\eta<\theta<\kappa\}$.

• By the compactness theorem, $\Gamma^*$ has a model - any finite subset of $\Gamma^*$ can only mention finitely many of the new constants, so an appropriate expansion of $\mathcal{A}$ will do the job.

• But any model of $\Gamma^*$ has cardinality $\ge\kappa$ - hence can't be $\mathcal{A}$ itself. But the reduct of such a model to the original language $\Sigma$ is a model of $\Gamma$, so there are $\Sigma$-structures satisfying $\Gamma$ which aren't (isomorphic to) $\mathcal{A}$.

(Note that the original formulation of PA was as a second-order theory; second-order logic can indeed capture $\mathbb{N}$ up to isomorphism.)

Answer 2

The answer is no. It is fairly easy to show that the axioms as stated are sufficient to characterize the natural numbers.

Proof of the following theorem suffices: $\forall_n\left[ n=1\lor \exists_m n=m^\prime\right].$ This is shown as follows: $P\left[1\right]$ is given, and $P\left[n\right]\implies P\left[n^\prime\right]$ is axiomatic.

But this really isn't the entire proposition which I wanted to demonstrate.

A more complete statement such as: every number is either $1$ or succeeds $1$ does appear to require second order logic. By $n$ succeeds $1$, I mean that the number $n$ is either $1^\prime$, or is the successor of a number which succeeds $1$.

$$S\left[x\right]\text{:=}\left(x=1'\lor \exists _y\left[S\left[y\right]\land x=y'\right]\right).$$

The proposition that $n$ is either $1$ or $n$ succeeds $1$ is now:

$$P\left[x\right]\text{:=}\left(x=1\lor S\left[x\right]\right).$$

The inclusion of the existential quantifier in $S$ raises this to logic of the second order.

Part of the reason for my confusion is that the authors of BBFSK seem to take the existence of the natural numbers as a-priori. This is especially true the in first chapter where they omit the first two of Peano's axioms in their second-order logic statement of them.