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

Question

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.

Answer 1

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.

Answer 2

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 $x_0$.

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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 $x \in N,~ x\neq 0.$