# Is it circular to include reachability from $0$ like this as a Peano axiom?

I am wondering whether it makes logical and semantic sense to include, as an axiom to define the natural numbers, that "every natural number is either $$0$$ or the result of potentially repeated successor operations from $$0$$ ", where one idea I have to define what it means to be repeatedly use the successor operations from $$0$$ is this:

If $$n= 0$$, we have repeatedly used the successor operation from $$0$$ $$0$$ times. If we have repeatedly used the successor operation from $$0$$ $$n$$ times to get $$k$$, then if $$k' = S(n)$$, we have repeatedly used the successor function $$S(n)$$ times to get $$k'$$.

Where $$n,k,k'$$ were any natural numbers and $$S$$ is the successor function. I am wondering because by using the term "repeated", we may be relying on a notion of number, as in order to define what it means to be repeated, we need to distinguish doing something once versus doing something twice; and as we are trying to define numbers, this therefore may introduce circularity into our definition.

However, is this really an issue? Every axiom is referring to the concept of a number, by stipulating something of natural numbers- saying $$0$$ is a natural number, or merely talking about $$n$$, is using a concept of a number in some way. So why can't we (potentially) refer to a number by using 'repeated'? Is my definition really different, inasmuch as potential circularity is concerned, than the induction axiom?

Anyways, if this is an issue, it seems this axiom can be logically and semantically salvaged if perhaps the word choice was to distinguish between doing something $$0$$ times, and doing something at all, whether it be done once or multiple times? If this statement doesn't work, what is another statement which is in the same "semantic spirit" (as in not merely logically equivalent, but a way to semantically express reachability from $$0$$ in a non-circular way)?

I am asking because in this post I asked whether this statement was equivalent to the axiom of induction. Here, I am concerned solely with the semantic nature and potential circularity of using this as an axiom, as opposed to in my other post how I was asking whether this statement is equivalent to the axiom of induction.

• How will you define repeated or reachable without using natural numbers? Like $\exists k \in \mathbb N ~ x = S^k(0)$ there you've just defined x in terms of k, but it is not really helping. Commented May 21 at 8:04
• The informal idea that the standard natural numbers are precisely the numbers reachable from 0 is accurate. But to accept a statement as an axiom in a way that can be studied formally (using model theory), you need to express it in a formal language that can be interpreted unambiguously (without issues around "word choice", conceptual circularity, etc.). For example, can you express reachability as a first-order sentence? Attempts to do this have resulted in the induction axiom.
– Karl
Commented May 21 at 19:31

I am wondering because by using the term "repeated", we are relying on a notion of number, as in order to define what it means to be repeated

This is exactly why formal mathematical language exists. Peano's axioms do not use the term "repeated", they merely state two rules:

1. $$0$$ is a natural number.
2. If $$n$$ is a natural number, then the successor of $$n$$, denoted as $$S(n)$$ (or $$n+1$$) is also a natural number.
• Those two axioms also apply to rational numbers. What O.P. said is strictly stronger. Commented May 21 at 7:59
• I edited my post to contain what I view to be a formal definition of repeatedness. What are your thoughts on it? Commented May 21 at 16:41
• Peano's axioms state more than just these two rules, regardless of formulation. Commented May 22 at 11:44
• @PrincessMia, as Karl points out, formal statements are not verbal, but written in the language of logic, to avoid ambiguity. The answer given by Dan Christensen gives an example of a formal statement, though it definitely isn't the one you're looking for (see the comment thread below the answer). Commented May 23 at 11:29

Let $$X, ~x_0$$ and $$f$$ be such that $$x_0 \in X$$ and $$f: X \to X$$. We can say that every element of $$X$$ (EDIT: with the possible exception of $$x_0$$) is accessible by a process of repeated succession starting at $$x_0$$ if

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

More formally:

$$\forall Y \subset X : [Y\neq \emptyset \land x_0 \notin Y \implies \exists a\in X : [a\notin Y \land f(a) \in Y]]$$

It can be shown that induction holds on $$(X, f, x_0)$$ with $$f: X \to X$$ being the successor function and $$x_0$$ being the "first element" of $$X$$ if and only if every element of $$X$$ accessible (in the sense given here) by a process of repeated succession starting at $$x_0$$.

EDIT:

You wrote:

... an axiom to define the natural numbers, that "every natural number is either 0 or the result of potentially repeated successor operations from 0.

The 5 usual Peano Axioms:

1. $$0\in N$$

2. $$\forall x\in N: S(x)\in N~~~~~~ (S: N \to N)$$

3. $$\forall x,~y \in N: [S(x)=S(y) \implies x=y]~~~~~~(S\textrm{ is injective})$$

4. $$\forall x \in N: \neg S(x)=0~~~~~~(S\textrm{ is not surjective})$$

5. $$\forall P\subset N: [0\in P ~\land ~\forall x\in P: S(x) \in P] \implies P = N~~~~~~(\textrm{Induction})$$

Your suggested axiomatization of the natural numbers might cover all but Axiom 3 here. The injectivity of $$S$$ is necessary to rule out the possibility that the successor function might "loop back" to some $$x \in N,~ x\neq 0.$$

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Commented May 24 at 15:32