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In chapter 2 of Terence Tao's Analysis, Tao goes over the axiomatic formulation of the natural numbers (Peano axioms). He explained the first four axioms:

Axiom 2.1. $0$ is a natural number.

Axiom 2.2. If $n$ is a natural number, then $S(n)$ is also a natural number.

Axiom 2.3. $0$ is not the successor of any natural number.

Axiom 2.4. Different natural numbers must have different successors.

Then he explains that this is not enough to formulate the familiar $\mathbb{N}$, because the set of "natural numbers" specified by only the first four axioms might contain all the elements we want, but it does not specify that those are the only elements of $\mathbb{N}$. (for example, $\{0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, \dots \}$ satisfies the four axioms but it contains more than $\mathbb{N}$.)

So we need a fifth axiom that can fill this role.

Axiom 2.5. (Principle of mathematical induction). Let $P(n)$ be any property pertaining to a natural number $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(S(n))$ is also true. Then $P(n)$ is true for every natural number $n$.

I am somewhat confused what this axiom tries to do. Am I right to think that this acts as an axiom which (1.) justifies the use of the logic of mathematical induction, and (2.) specifies precisely the elements of the set of natural numbers, both at the same time?

If so, then isn't (1.) a bit too overkill, if we are just trying to formulate the set of numbers isomorphic to $\mathbb{N}$?

Edit: To clarify, the confusion arises from the idea that axiomatizations should be as simple as possible. The principle of mathematical induction seems like a heavy axiom just to state that "every natural number is either $0$ or a successor of $0$".

Edit: This question does not answer my confusion. The question in the link asks why the induction axiom "only allows the natural numbers as we know them", whereas my confusion comes from the fact that induction seems "too much" just to state the fact that the natural numbers consists of $0$ and repeated successors of $0$.

Dreme
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  • If you have the naturals, why wouldn't you want to get induction at the same time? In other words, it's not clear from the Question what the objection/problem/difficulty is. – Eric Towers Dec 26 '24 at 16:23
  • @EricTowers If the motivation was to build an axiomatic system to get $\mathbb{N}$ , I thought the goal was to do it in as simple axioms as possible. So we can do something like "Every natural number is either 0 or some repeated successor of 0" without mentioning mathematical induction. Sure we would want to get induction, but isn't the whole point of axiomization, to do it in fewest, simplest possible statements? – Dreme Dec 26 '24 at 16:33
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    @Dreme How do you express "is a repeated successor of 0"? – spaceisdarkgreen Dec 26 '24 at 16:50
  • @spaceisdarkgreen Well.. I couldn't think of a better way to formally express that sentence. Is this how we settled to this conclusion, that induction is the simplest way to state that sentence? It just feels a little weird, to use a whole axiom schema just to state that fact... (This also comes from the assumption that axiom schemas are more complex than simple axioms) – Dreme Dec 26 '24 at 16:53
  • @Dreme It's not an axiom schema, it is a second order axiom (a first order schema can't even express that... there are always nonstandard models). The most direct way to express it would be as an infinite disjunction in infinitary logic, but infinitary logic is less familiar and generally viewed (perhaps unfairly) as less natural than second order logic. – spaceisdarkgreen Dec 26 '24 at 16:58
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    Another tack: One wants to make an infinite number of "this thing is a natural number" statements. If your logic is finite, no proof can get you all the naturals. To get all the naturals at once, you need some way to get infinitely many statements at once from a finite specification/proof. – Eric Towers Dec 26 '24 at 17:12
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    So what would you suggest as an alternative? You suggest "So we can do something like "Every natural number is either 0 or some repeated successor of 0" without mentioning mathematical induction." How on earth can you define "REPEATED" without mentioning something equivalent to induction? What does "repeated" mean? To my mind induction is the only way to say "repeated" and induction says nothing more or less than "through repeated iteration we get everything and everything we get is all we want" – fleablood Dec 26 '24 at 17:40
  • @EricTowers Putting it that way makes much more sense why some sort of induction may be needed. Thanks for the input! – Dreme Dec 26 '24 at 18:05

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As you pointed out in the question, the first four axioms ensure that $0,1,2,\dots$ are natural numbers, but they leave open the possibility that some additional, unwanted things, like $0.5,1.5,\dots$ are also natural numbers. (There are even worse-looking examples, like the set of all real numbers except the negative integers.) We need another axiom to exclude such unwanted stuff.

If you look more closely, you see that all the bad examples that satisfy the first four axioms contain the genuine natural numbers along with more elements. (That's because the first four axioms already ensure that all the genuine natural numbers will be in the bad examples.) So the genuine natural numbers are the smallest example; that is, if I write $\mathbb N$ for the set of genuine natural numbers and write $E$ for any example satisfying axioms 1--4, then $\mathbb N\subseteq E$.

That gives us a clue about prohibiting all the bad examples and thereby describing exactly $\mathbb N$: We want the smallest of all examples. That is, any $E$ that satisfies axioms 1--4 must be a superset of $\mathbb N$.

That's essentially the axiom we need, but it can be simplified a bit. We can equivalently say that any $E$ that is a subset of $\mathbb N$ and satisfies axioms 1 & 2 is $\mathbb N$. The point is that, being a subset of $\mathbb N$, such an $E$ will automatically satisfy axioms 3 & 4. So we can formulate our new axiom as:

If $E\subseteq\mathbb N$ and $0\in E$ (i.e., $E$ satisfies axiom 1) and whenever $n\in E$ then also $S(n)\in E$ (i.e., $E$ satisfies axiom 2), then $E=\mathbb N$.

That's essentially the induction axiom that you quoted. The only difference is that the quoted axiom talks about properties $P$ of natural numbers whereas my version talks about subsets of $\mathbb N$. But those are essentially the same thing. Any property $P$ gives you a set $E=\{n\in\mathbb N:P(n)\}$, and any set $E$ gives you a property $P(n)\iff n\in E$.

Finally, let me point out that, in this whole discussion, the goal was merely to completely describe $\mathbb N$ by prohibiting all the bad examples. I was not "aiming" to get an induction principle. But I got one anyway. What luck! Well, actually, it's not luck. In fact the induction principle is saying exactly that we're dealing not with a bad example but the genuine $\mathbb N$. The induction principle is just a formalization of the intuition that a natural number needs to be reachable from $0$ by (finitely often) repeated use of the successor function.

Andreas Blass
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  • We could define the natural numbers by just the four first axioms and saying that $\mathbb N$ is the smallest set satisfying the axioms, and we could define arithmetics on such a set, but we wouldn't be able to show any theorems like commutativity or associativity. – md2perpe Dec 26 '24 at 21:50
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    @md2perpe Saying that $\mathbb N$ is the smallest set satisfying the first four axioms is just what the induction principle says. – Andreas Blass Dec 26 '24 at 21:58