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The natural numbers are defined recursively as

$$\mathbb{N} = \{x\in \mathbb{N} \,|\, x = 0 \lor (\exists y\in\mathbb{N})[S(y) = x]\}.$$

Semantically, this means that $\mathbb{N}$ contains $0$ and that $\mathbb{N}$ is closed under the successor function. Similarly, many theorems on the naturals (such as commutativity and associativity of addition and multiplication) can be proved using induction; that is for any function $f$(defined set-theoretically as here) whose domain is $\mathbb{N}$ and whose codomain is $\mathbb{B}$ (the set containing "true" and "false", however they may be represented), we can say that

$$f(0) \land (\forall x \in \mathbb{N})[f(x)\implies f(S(x))]\implies(\forall x \in \mathbb{N})[f(x)].$$

Semantically, this states that if $0$ satisfies some function, and if a number satisfying the function implies that its successor satisfies the function, then the function is true for all natural numbers. My question is, is there a way to prove the validity of general induction proofs on the naturals using the definition of the naturals (and if so, can a concrete example be given)? If axioms are required, what are the weakest axiom(s) that are necessary to prove the validity of general induction proofs (already assuming the basic set theory axioms and definitions)?

Sorry if I'm misunderstanding something about logic or the construction of the naturals -- I'm just a novice trying to gain basic understanding.

Thanks for your guidance!

Cameron Buie
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Mathemagician314
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    Well, set theory tends to assumes there is an ordinal, and defining the natural numbers is easy, but it is essentially defined so that induction follows. So set theory needs an axioms which are specifically there to define the natural numbers, where it is easy to prove induction is true. But that axiom is not induction. However, the axiom is somewhat artificial to give us induction. – Thomas Andrews Dec 18 '24 at 01:56

4 Answers4

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The natural numbers are defined recursively as

$$\mathbb{N} = \{x\in \mathbb{N} \,|\, x = 0 \lor (\exists y\in\mathbb{N})[S(y) = x]\}.$$

This doesn't work.

Semantically, this means that $\mathbb{N}$ contains $0$ and that $\mathbb{N}$ is closed under the successor function.

Note that $\mathbb{N} = \{0\}$ and $S(0)=0$ would satisfy this condition.

My question is, is there a way to prove the validity of general induction proofs on the naturals using the definition of the naturals?

Peano's Axioms can be thought of as defining the natural numbers. They include the Principle of Mathematical Induction. Each of these axioms can, however, be derived if you accept that there exists even a single Dedekind infinite set. There will exist a subset of that infinite set on which each of Peano's Axioms will hold.

Note that induction can hold on any set $X$ (possibly finite) with $f: X \to X$ (the successor function) and $x0\in X$ such that:

$X= \{x_0, f(x0), f(f(x_0)), \cdots \}$

In words, every element of $X$ can be reached by a process of repeated succession starting at $x_0$.

Example: $X=\{0, ~1\}$, $f(0)=1$ and $f(1)=0$

Induction holds on $(X,f,0)$ since we can show that:

$\forall P\subset X: [0\in P ~\land~ \forall a\in P: f(a)\in P \implies P=X]$

Hint: There are only 4 subsets of $X$ to consider.

Dan Christensen
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It can be proved by the well ordering principle that proof by induction always works. Assume $P(n)$ is a statement such that $P(0)$ is true and such that if $P(n)$ is true, then $P(n+1)$ is true. If $P(n)$ were not true for all $n$, there must be a smallest value of $n\ge1$ for which $P(n)$ were false, by the well ordering principle. But then consider $P(n-1)$…

Alex Ortiz
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  • Thank you! This makes a lot of sense. As a further question, is it possible to write this out as a formal proof using the notation of first-order logic? – Mathemagician314 Dec 18 '24 at 02:10
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    @Mathemagician314: In principle yes. You can explore a fully formal proof in Metamath notation which concludes induction from the ZF axioms: that's fairly close to using first-order logic + ZF axioms, but a bit more convenient due to abbreviation mechanism. The conclusion is line 37. As you'll see, fully formal proofs are looong. – Z. A. K. Dec 18 '24 at 05:24
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    This is just shifting the problem from one place to another. The well-ordering principle for the natural numbers is not an axiom either. It needs a proof, too. And the usual way to prove it is by induction. As Thomas Andrews pointed out in a comment, the magic is in the definition of the natural numbers. By definition, they are basically the one and only set where induction works. The existence of such a set must be ensured by an axiom (the axiom of infinity), but induction itself is not one. – Tzimmo Dec 18 '24 at 09:24
  • @Tzimmo Induction works over any Artinian set (of course, in a slightly more general formulation but nevertheless similar in spirit). – ΑΘΩ Dec 19 '24 at 14:08
  • Hi Alex. Apologies, but I have downvoted this answer for the following reason: as discussed in the comments to this answer, and the linked paper, it is not accurate to say the well-ordering principle implies the induction principle. More specifically, there are models of second-order arithmetic satisfying the first four Peano axioms, and the well-ordering principle, but not the induction principle. – Joe Dec 22 '24 at 20:37
  • In addition, in most (all?) foundations of mathematics that are commonly used, like say ZFC and ETCS, the well-ordering principle is not an axiom. It requires proof. It's only an "axiom" in the same way that the group axioms are "axioms". – Joe Dec 22 '24 at 20:38
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The axiom of infinity tells us that that there exists an "inductive set", i.e.: $$\exists X. \emptyset \in X \land \forall x\in X. S(x) \in X$$

$\mathbb{N}$ is defined as the intersection of all inductive sets (which can be easily proved to also be an inductive set).

The definition of $\mathbb{N}$ in the original question is not really a definition in the usual sense. You're using $\mathbb{N}$ in the body of your definition! That's a circular definition (at least when interpreted literally). You can't do that! So I'll use my definition instead in what follows.

To prove induction, take some property of natural numbers $P$ such that $P(0)$ and $\forall n\in \mathbb{N}. P(n) \Rightarrow P(S(n))$. Now take: $$Q = \{n\in \mathbb{N}: P(n)\}$$ The assumptions about $P$ mean that $Q$ is an inductive set, which by definition of $\mathbb{N}$ means that $\mathbb{N} \subseteq Q$ (and thus $Q = \mathbb{N}$). Therefore: $$\forall n\in \mathbb{N}. P(n)$$ QED

Tomek Czajka
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    An alternative, more intuitive axiom of infinity might be: $\exists X,f,x_0:[f:X\to X~ \land~ x_0 \in X ~\land~ \forall a,b,c \in X: [f(a)=b~ \land~ f(c)=b \implies a=c]~\land ~ \forall a\in X: f(a)\neq x_0]$. – Dan Christensen Dec 19 '24 at 00:00
  • In other words, there exists a Dedekind infinite set. From this axiom, we can easily construct a subset on which Peano's Axioms hold. – Dan Christensen Dec 19 '24 at 00:06
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You need the Peano axioms. 1. 0 is a natural number. 2. If n is a natural number then n+1 is a natural number. 3. If a set meets conditions 1. and 2. then it contains all natural numbers.

Now take a predicate A(n) with the property that A(0) is true and A(n) implies A(n+1). Let S be the set of numbers n for which A(n) is true. Then the peano axioms immediately show that S contains all natural numbers.

So you need axioms for complete induction but it is a quite trivial step.

gnasher729
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