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My title probably doesn't explain my worry / concern too well, but it's the best title I could think of. I am researching construction of real numbers for a college project, and as a consequence I am researching the Peano axioms. A couple of sources I have read come to the conclusion that $$ \{0,1,2,3,...\} \subseteq \mathbb{N} $$ (where ${1=S(0)}$, ${2=S(1)}$,...) and by the axiom of induction we have $$ \mathbb{N} \subseteq \{0,1,2,3,...\} $$ hence $$ \mathbb{N}=\{0,1,2,3,...\} $$ (in other words, the Natural Numbers are the set of all possible successors of $0$). The part that makes me feel uneasy is the definition of ${\{0,1,2,3,...\}}$. Doesn't defining this set in the first place use induction in some sense? Like - we start with $0$. Then ${S(0)}$ cannot be $0$, and so we define ${1:=S(0)}$. Then ${S(1)}$ can be neither $0$ or $1$, and so we define ${2:=S(1)}$... I'm not sure how doing this for finitely many elements justifies the existence of the set ${\{0,1,2,3,...\}}$. Does what I'm saying even make sense? Am I overthinking things? Thank you!

EDIT: I should specify, I am following the definition of the axiom of induction as is on Wikipedia (https://en.wikipedia.org/wiki/Peano_axioms)

EDIT 2: source example: http://www2.hawaii.edu/~robertop/Courses/TMP/7_Peano_Axioms.pdf , page $3$

Riemann'sPointyNose
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    "Prove $\langle \mathbb{N}, x \mapsto x +1, 1 \rangle$ is a Peano system without circular reasoning" is essentially the same kind of question, and most likely the answer there should address your question. You're not over-thinking this because indeed you cannot construct $ℕ$ without using something that is essentially as powerful as induction to begin with. More details in the linked post. – user21820 Jan 31 '21 at 17:32
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    Note that Dan's post on the linked thread is completely bogus, as any professional logician can confirm; it is simply impossible to non-circularly construct an inductive structure out from nothing. – user21820 Jan 31 '21 at 17:38
  • @user21820 I see yeah! Thank you for leaving a comment and linking the thread. In retrospect, I believe my whole confusion came down to taking the "pseudo proof" in the linked article too literally. I guess trying to show Peano axioms imply ${\mathbb{N} = {0,1,2,3,...}}$ wouldn't make sense, because in order to do that you need to formalise a definition for ${{0,1,2,3,...}}$ in the first place, which is what Peano axioms are trying to do. It's like trying to "prove" epsilon delta limit definition is the "same" as what we mean intuitively by limit - it doesn't make sense – Riemann'sPointyNose Jan 31 '21 at 20:57
  • Absolutely right! =) – user21820 Feb 01 '21 at 06:03

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The notation $\{0,1,2,3,\cdots\}$ is being used here to symbolize "A set containing $0$, containing $S(n)$ whenever it contains $n$, and containing nothing else." It is true that the Peano axioms do not justify the existence of this set, but that is no surprise, since the Peano axioms do not deal with sets at all! With this line of thought, your two claims become "For every $n\in \mathbb N$, we have $S(n)\in \mathbb N$" and "If a set contains $0$ and $S(n)$ for every $n$ which it contains, then it contains $\mathbb N$", both of which are directly asserted by the Peano axioms. So your sources, at least as you paraphrase them, are saying nothing at all.

Kevin Carlson
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  • Ah I see, so what you are saying is that ${{0,1,2,3,...}}$ is just shorthand for the inductive set? – Riemann'sPointyNose Dec 14 '20 at 23:44
  • @Riemann'sPointyNose Yeah, exactly, I see no other way to interpret it consistent with looking at Peano arithmetic in its own right, rather than as something to be interpreted within an encompassing set theory. – Kevin Carlson Dec 15 '20 at 03:05
  • Thank you! I gave you the solution checkmark since you did technically answer my first question, I really appreciate it. I do now have a separate question I believe so I will post this as a separate question (since it's not the same question technically I guess). Thanks again! – Riemann'sPointyNose Dec 15 '20 at 17:03