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In this question, and specifically in this answer, it is explained that $\mathbb{N}$ is not definable in $\mathbb{R}$ using the language of ordered fields (using $0$, $1$, $+$, $\cdot$, $<$) with a first-order formula. In the answer there are some words about second-order logic, but they're very brief.

So I'm wondering:

  1. Is it possible to use a second-order formula to define $\mathbb{N}$? What would it look like?
  2. Is it possible to add an axiom schema in first-order logic (adding, effectively, infinite many axioms) to define $\mathbb{N}$? What would it look like?

(I'm interested in both questions)

The same answer shows that a second-order formula like $$\forall \phi \big[\phi(0)\land\forall x\big(\phi(x)\rightarrow\phi(x+1)\big)\big]$$ defines an inductive set (although the formula there uses set notation), but I don't know if it's enough since it's still necessary to show that there is a "minimum inductive set", isn't?

Daniel
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1 Answers1

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$\mathbb N$ is the set of all $y\in\mathbb R$ such that $$ \forall\phi\,\big([\phi(0)\land\forall x\,(\phi(x)\to\phi(x+1))]\to\phi(y)\big); $$ in other words, every inductive set contains $y$.

Andreas Blass
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  • I'm not very familiar with second-order logic. Does this formula categorize $\mathbb N$ as a "minimum inductive set"? (With minimum I mean, of course, that is a subset of every inductive set) In FOL usually you prove that such a set exists, that is really a subset of every inductive set and then you denote that set as $\mathbb{N}$ or $\omega$ (depending on the theory). Also, do you have an answer to second question? I'm interested in both of them. – Daniel Sep 13 '19 at 15:00
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    By the upward Löwenheim-Skolem-Tarski theorem, no collection of first-order statements can uniquely (up to isomorphism) characterize $\mathbb N$ (or any other infinite structure). – Andreas Blass Sep 13 '19 at 20:26
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    @Daniel Showing that the definition works is different from giving the definition itself. This does define $\mathbb{N}$ as the intersection of all inductive sets; properties of this definition (e.g. that said intersection is nonempty) do indeed require verification if we want to use them, but this does define $\mathbb{N}$ in $\mathbb{R}$. – Noah Schweber Sep 18 '19 at 19:35