Definition of Natural Numbers which gives rigorous formulation to "principle of induction"?

Question

So I know the first order theory that is Peano Arithmetic (just to ensure we are on the same page, it is a first order theory with signature $\{0, S, +, \cdot, = \}$ with the axioms: (i) All axioms from theory of pure equality, (ii) Zero and Successor Axiom (iii) 2 axioms on defining $+$ (iv) 2 axioms on defining $\cdot$ and (v) Induction (axiom schemata).

But how are the natural numbers defined?

I was thinking it might be some model of the above first order theory ... but I was unable to figure out how to formally define the domain of discourse and the mapping to each predicate and function symbol?

Seeing the comments, it seems the set-theoretic definition of $\mathbb{N}$ is the way to go, but could you give a proof that this set satisfies the induction principle. You could take of from the answer mentioned here: Set theoretic construction of the natural numbers

Edit: As mentioned in comments, the link above gives a proof of induction for natural numbers - it is item 5 in the theorem labelled "Peano Axioms".

Answer 1

If you are working in first order logic with the Peano axioms there is no definition of the naturals. Induction is one of the Peano axioms. The naturals can be any set that satisfies the Peano axioms. You can assure that all the naturals you know are there because you can reach them with the successor function. You would like to say that there aren't any more naturals, but you can't do that within first order logic. In fact there are continuum many countable models of all the PA axioms.

Alternately you can define a set as the naturals. It will presumably be inductive, so you have that for free. You can then study the properties of the naturals in this set. Now you have to prove that the Peano axioms are true for the set you have chosen if you want to use them.

Answer 2

Proving The Induction Schema for Natural Numbers looks like this:

Meta-Theorem

Let $\phi$(x) be a well-formed formula of the Language of Set Theory

Then,

[$\phi$(0) $\wedge$ $\forall$n$\in$N($\phi$(n) $\rightarrow$ $\phi$(n+1))] $\rightarrow$ $\forall$n$\in$N($\phi$(n))

Proof:

Assume $\phi$(0) $\wedge$ ($\forall$n$\in$N($\phi$(n) $\rightarrow$ $\phi$(n+1))

(RAA) Assume $\exists$n$\in$N($\neg $$\phi$(n))

So, there is a counter-example to $\phi$(n) holding for every n

Consider the Set of Counter Examples,

A = { n$\in$N| $\neg $$\phi$(n) } is Nonempty by our Assumption.

As the Natural Numbers are well-orderd, Consider the <-least counter-example

We Denote it by x, so $\neg $$\phi$(x)

Notice that by assumption $\phi$(0) holds, so x ≠ 0.

As every natural number is either 0 or a successor of a natural number ( Peano Axiom ( or by construction, in Set Theory)

We have that x = y + 1 for some natural number y.

But, by assumption

$\forall$n$\in$N($\phi$(n) $\rightarrow$ $\phi$(n+1))

Specialize for n = y, so

$\phi$(y) $\rightarrow$ $\phi$(y+1)

As x is the least counter-example and y < x, y is not a counter-example so $\phi$(y)

By Modus Ponens

$\phi$(y+1), but y+1 = x

So, $\phi$(x) this contradicts x being a counter-example. So, our assumption that $\phi$(x) has a counter-example has been reduced to a contradiction.

Thus, there are no counter-examples.

Answer 3

In almost any development of mathematics — whether in set theory, or another formal system (e.g. type theory), or just in ordinary human language — the natural numbers $\newcommand{\N}{\mathbb{N}}\N$ themselves are one of the first, most fundamental objects defined. Definitions of logical syntax, models, and PA all come later, and don’t affect that original definition of $\N$ — so the naturals really are the specific set, not “any model of PA”.

For perspective: A very common mistake, for relative newcomers to logic, is to over-weight the importance of PA. I think a useful comparison is semiring structure. You can prove “$(\N,+,\cdot)$ is a semiring”, and this suffices to prove several facts about $\N$ — but there are many other different semirings, and many properties of $\N$ that don’t follow just from the semiring axioms, so it’s clear that these axioms are just a partial description of $\N$, an axiomatisation of some of the structure it carries — not a complete description of $\N$, and certainly not a definition of it. PA goes much further down the same road — it describes many more properties that $\N$ satisfies, and is strong enough to let us express and prove many more interesting things about $\N$ than we can from the semiring axioms, but it’s still not a complete description of $\N$, let alone a definition.

With that perspective in mind, the fact that $\N$ satisfies PA is just another theorem we need to prove. How we prove it depends on exactly which definition we use. A common approach approach in ZF or similar set theories is as follows: define 0 as the empty set, define a “successor” operation on sets by $s(x) = x \cup \{x\}$; then define $\N$ as $$\{ x \mid \text{every set containing $0$ and closed under $s$ contains $x$}\};$$ spelled out in more primitive terms, it’s the unique set $S$ such that for any set $x$, $x \in S$ if and only if for every set $Y$, if $Y$ contains $0$ and is closed under $s$, then $x \in S$. (The axiom of extensionality tells us there’s at most one such set $S$; the axiom of infinity implies that some such $S$ exists.) This definition immediately implies “for every $Y$, if $Y$ is closed under $0$ and $s$, then $\N \subseteq Y$”, and is often phrased as “$\N$ is the least set containing $0$ and closed under $s$”.

From this definition, we can now prove the principle of induction quite easily. For any property $P(x)$ of natural numbers, if $P(0)$ holds and $P(x) \Rightarrow P(s(x))$ for all $x \in \N$, then the set $Y_P := \{ x \in \N \mid P(x) \}$ contains $0$ and is closed under successor, so is all of $\N$; that is, every $x \in \N$ satisfies $P(x)$.

It’s worth noticing that this is a stronger form of induction than the form required to show $\N$ models PA. This full version says that induction holds for any property of $\N$. The PA scheme just says induction holds for properties expressed using first-order logic and $0,1,+,\cdot$ — which covers many properties, but not all. In any non-standard model $(M,0_M,s_M)$ of PA, the full principle of induction will fail: the property “$x \in M$ is standard”, i.e. “$x$ is of the form $s_M^n(0_M)$, for some $n \in \N$”, will hold for $0_M$ and be closed under $s_M$, but doesn’t hold for all elements of $M$, since $M$ is nonstandard.