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I am an engineer who is learning real analysis myself. I read Peano axioms which are about the set of natural numbers $\mathbb{N}$. However, I find the statements, especially the word "successor" is not well defined, and cannot be converted into well-formulated formulae in first-order logic. Could anyone explain how to accept the word "successor"?

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  • $\begingroup$ The statement is that there is a function $S:\mathbb N \to \mathbb N$ with the properties set out in the axioms. It is labelled the successor function, and its application $S(n)$ is described as the successor of $n$ $\endgroup$
    – Henry
    Nov 25, 2020 at 23:17
  • $\begingroup$ Not sure what to tell you. Axioms by definition are not "rigorous" or "provable". They are nothing more than arbitrary rules, and sometimes backed by nothing more than gut, intuition, instinct, or emotional reactionism to contrary statements. We dont allow contradiction in math... why? Cant prove why without resorting to putting the cart before the horse. At the end of the day it just "feels right". $\endgroup$ Nov 25, 2020 at 23:49
  • $\begingroup$ In the case of Peanos axioms, they were deliberately constructed to be as few as possible and as complete as possible to describe the construction of the Naturals. They dont have to be the only set of axioms that do this, and there is some controversy behind some axioms of math, but if it works and no contradictions fall out of it, we are content to use them. $\endgroup$ Nov 25, 2020 at 23:51

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In the context of Peano, we use the language of first-order logic with equality, enhanced by two symbols: $0$ and $S$, which are intended to stand for the number zero and for the successor operation. It is the addition of these symbols to the language that makes the respective axioms (e.g., $\forall n\colon \neg(Sn=0)$ well-formed. There is nothing to be defined about $S$ (or $0$). The meaning of these symbols is comprised by the axioms.

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    $\begingroup$ Given the focus on first-order logic this isn't really correct: $\{0,S\}$ is the language of second-order Peano arithmetic. First-order $\mathsf{PA}$ has to use addition and multiplication as well. $\endgroup$ Nov 25, 2020 at 23:53
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I am a physicist, so my perspective is a not quite that of either an eningeer or a mathematician. As usual, my empathy lies with the engineer, but my aim is toward the mathematician.

The 5 Peano axioms at the beginning of Kenneth Ross's excellent text $\it{Elementary~ Analysis: The Theory~of~Calculus}$ could/should be called (for teaching purposes) the Successor Axioms. This is because the notion of a successor is foundational to them. I would define a successor operator $\texttt{++}$ with notation that evokes computer science, and organize axioms differently. To be clear, the successor of $\texttt{n}$ is denoted $\texttt{++n}$ and has the value $\texttt{n+1}$. (So somewhere along the line, we have learned what "add 1" means.) Then, as Ross states, the natural numbers $\mathbb{N}=\{1, 2, \dots \}$ have the following "properties" (which we treat as axioms)

  1. If $\tt{n}$ is an element of $\mathbb{N}$, then its successor $\texttt{++n}$ is an element of $\mathbb{N}$. (Ross, Peano N2)
    • a) 1 is an element(Ross, Peano N1), but
    • b) 1 is not the successor of any element of $\mathbb{N}$. (Ross, Peano N3)
  1. If $\texttt{++m = ++n}$, then $m=n$. (There is no convergence of separate branches of successors in $\mathbb{N}$; it is one long string.) (Ross, Peano N4.)
  2. Any subset of $\mathbb{N}$ that contains 1 and contains $\texttt{++n}$ if it contains $\texttt{n}$ constitutes the entire set $\mathbb{N}$. (Ross, Peano N5)

So I have four axioms, with the second one being a joining of two of the Ross Peano axioms. Ross Peano N1 and N3 as separate axioms may be logically indicated, but their recombination is warranted, I believe, since each axiom should be a separate statement about a successor property of $\mathbb{N}.$ Separating them disguises the centrality of successors to the axiomatic basis of $\mathbb{N}.$

Perhaps this appeals more to an engineer (and perhaps also to a computer scientist?) without distressing mathematicians too, too much. (?)

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