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I'm starting to study mathematical logic, concretely I'm interested in how PA can prove theorems about infinite ordinals, and I came with this very naive question about induction over infinite ordinals.

It is known (Gentzen 1943) that induction up to every ordinal $\alpha<\epsilon_0$ is provable under PA. At the same time, the proof theoretic strength of PA is precisely $\epsilon_0$, which implies that $\epsilon_0$ is the least ordinal $\alpha$ for which $\alpha-$induction is not provable. So, by Gödel’s incompletness theorem, $\epsilon_0-$induction is not provable in PA.

However, if one understands $\alpha-$induction as the fact that $\alpha$ is well-ordered, and the definition of $\epsilon_0$ as $$ \epsilon_0=\sup\{\omega,\omega^{\omega},\omega^{\omega^\omega},\ldots \} $$ is difficult for me to see why the implication of the question's title is not true.

My intuition is that for a property to hold for every $\omega^\alpha$ does not imply that the property holds for $\epsilon_0$ for the same reason that one can write a property that holds for finite ordinals but fails in the limit ordinal $\omega$, but I can't convince myself about this reasoning.

Does anyone have any alternative explanation?

Tankut Beygu
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Somerandommathematician
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  • By the way, the property I was thinking of is for example "every element is a succesor". This is true for all $n\in\mathbb{N}$, but not for the limit ordinal $\omega$. – Somerandommathematician Oct 23 '24 at 19:05

1 Answers1

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Below I'll let $\mathsf{IND}(\alpha)$ be the specific version of induction relevant to Gentzen, and I'll conflate an ordinal $\alpha\le\epsilon_0$ with a "standard notation" for said ordinal.

The statement $$\forall\alpha<\epsilon_0[\mathsf{IND}(\alpha)]\quad\implies\quad\mathsf{IND}(\epsilon_0)$$ is indeed true. In fact, it can even be proved in $\mathsf{PA}$! Moreover, for each specific $\alpha<\epsilon_0$, we do indeed have that $\mathsf{PA}$ proves $\mathsf{IND}(\alpha)$.

However, we do not have in $\mathsf{PA}$ a uniform proof of the above which works for every $\alpha<\epsilon_0$. That is, while "For all $\alpha<\epsilon_0$, $\mathsf{PA}$ proves $\mathsf{IND}(\alpha)$" is true, the statement "$\mathsf{PA}$ proves [that] for all $\alpha<\epsilon_0$ [we have] $\mathsf{IND(\alpha)}$" is false. And so there's no way to leverage the (correct!) fact above, within $\mathsf{PA}$ alone.

Noah Schweber
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  • I see, so the key point is that PA can't prove $\forall\alpha(\textbf{IND}(\alpha))$ (I'm guessing that this is because that statement is a second order sentence). – Somerandommathematician Oct 24 '24 at 07:06
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    @Somerandommathematician Nope, because (if we look carefully under the hood so to speak) we're talking about ordinal notations instead of ordinals as such, this is all nicely first-order. The problem is that, as $\alpha$ increases, the PA-proofs of $\mathsf{IND}(\alpha)$ get trickier and trickier in a precise sense, so - within PA alone - there is no way to amalgamate them into a single proof. This is similar to how for each specific $n$, PA proves "$n$ is not (the code of) a proof of $0=1$ from PA," but (by Godel's 2nd) PA does not prove "no $n$ is (the code of) a proof of $0=1$ from PA." – Noah Schweber Oct 24 '24 at 22:46
  • Ok, so i think that I must read Gentzen's original proof of $\mathsf{IND}(\alpha)$. Do yo know where can I find the traslation? Thanks a lot btw. – Somerandommathematician Oct 25 '24 at 07:23