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From Terence Tao's Analysis I, Axiom 2.5 for the natural numbers reads

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My intuition behind this axiom is that every natural number is an element of a "chain" of natural numbers that goes back to $0$ and is built by successively applying the successor function to elements of it. This axiom then prohibits the existence of a "parallel chain" that runs outside of the "main" one, which contains $0$.

My question is: to prove that this axiom does indeed imply the existence of only this, one, correct, "chain" that starts at $0$, must we produce a specific property that shows this to be the case, or is it possible to prove this implication without alluding to any specific property, by directly using the fact that any property must satisfy the axiom? My question isn't whether we can find such a property; it might even be a very immediate one, but whether we must explicitly invoke it to prove the implication.

My mind keeps going back and forth on this one, and I can't pinpoint why.

jvf
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    The axiom does not characterize the natural numbers. You need to add in the axiom asserting that $n{++} \neq 0$. Without that natural numbers taken modulo $N$ satisfy the induction principle for any $N$. To show that these two axiom characterise the reals, you need to be clear about what logical system you are working in, e.g., first-order set theory of higher-order logic. – Rob Arthan Feb 05 '24 at 21:25
  • @RobArthan I didn't mean to convey that that axiom alone characterizes the natural numbers, my question was just focused on that one specifically. – jvf Feb 05 '24 at 22:30
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    Well you wrote "this axiom does indeed imply the existence of only this, one, correct ..."? So what are you actually asking? As I said in my previous comment, you need to be clear about the logical system you are working in. – Rob Arthan Feb 05 '24 at 22:34
  • @RobArthan I'm not knowledgeable in logical systems, so I truly don't know. What I meant with "only" was that >only< those numbers that can be accessed with sufficient applications of the successor function to $0$ are the only natural numbers. – jvf Feb 06 '24 at 02:45
  • For this and more see the award winning article by Henkin cited here. – Bill Dubuque Feb 27 '24 at 04:59

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The Peano axioms include the induction axiom but are not sufficient to prove that there are no numbers outside the set you can access from $0$ with the successor function. The induction guarantees that any statement true about all the naturals you can access with the successor function, which we call standard naturals, is true about all the numbers. In fact, the nonstandard naturals, if they exist, are greater than any standard natural, so any statement true about all large enough standard naturals is also true about all the nonstandard ones.

Ross Millikan
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  • But doesn't a property $P(n)$ expressed as "$n$ is either $0$ or can be accessed from $0$ with sufficient applications of the successor function" guarantee that those elements that can be accessed from $0$ with the successor function are the only natural numbers? The book makes a point of saying that this axiom ensures that those are the only natural numbers, so that's confusing. Otherwise, why is this even an axiom? – jvf Feb 06 '24 at 02:38
  • You can't define that property within the language of PA. The question is whether you are working in first order logic, which only allows you to talk of numbers but not sets of numbers or second order logic, which allows sets of numbers. PA is first order logic, which has very nice proof properties. It also has the compactness theorem, which says that if every finite subset of a set of statements has a model the whole set (which may be infinite) has a model. You can use this to show there must be models with nonstandard elements. – Ross Millikan Feb 06 '24 at 03:02
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I'm not sure what you mean by a "parallel chain." But I often find it useful to think of induction using the well ordering principle. Suppose you have some property $P$ that satisfies Axiom $2.5$. Let's also assume (toward a contradiction) that there's some natural number $n$ for which $P(n)$ fails. Then the set $C=\{ k \in \Bbb N \mid \lnot P(k) \}$ would be a non-empty set of natural numbers.

Since $C \neq \varnothing$, the well ordering principle tells us that $C$ would therefore have to have a least element $k$. We know that $k$ can't be $0$ because we're told that $P(0)$ holds. Therefore, $k=n+1$ for some $n \in \Bbb N$. But our assumption that $k=n+1$ is the least element of $C$ means that $n \notin C$, which in turn means that $P(n)$ holds. But if $P(n)$ holds, then Axiom $2.5$ tells us that $P(k)$ (which is $P(n+1)$) also holds, so $k=n+1$ isn't a member of $C$ after all. That's a contradiction.

Our initial assumption that $C$ is non-empty must therefore be false, so $P(n)$ must hold for all natural numbers $n$.

Robert Shore
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  • Given other axioms, the well-ordering principle slyly slides in the additional axiom $n{++} \neq 0$ that I mentioned in my comment. I don't think this is the best way to think about induction as a principle in its own right. – Rob Arthan Feb 05 '24 at 22:31
  • @RobArthan I think any reasonable set of axioms that establishes an order respecting the successor operation forces $n++ \neq 0$. – Robert Shore Feb 05 '24 at 22:48
  • The induction principle itself does not require $n{++} = 0$. It is true in $\Bbb{Z}_n$ for any $n$. As I say, I don't think your suggestion is the best way to think about the induction principle in its own right". – Rob Arthan Feb 05 '24 at 23:00