Let $\mathsf{PA}_{\Sigma^1_1}$ be the theory in second-order logic gotten by extending the usual first-order Peano axioms to include arbitrary $\Sigma^1_1$ formulas in the induction scheme. My question is:

Does $\mathsf{PA}_{\Sigma^1_1}$ have any nonstandard models?

Note that a model of $\mathsf{PA}_{\Sigma^1_1}$ is exactly a model of $\mathsf{PA}$ with no (nontrivial proper) $\Sigma^1_1$-definable cuts.

If we replace $\Sigma^1_1$ with $\Pi^1_1$ the answer is immediately negative, since the set of standard elements of a model of $\mathsf{PA}$ is $\Pi^1_1$. However, nothing similar seems to work for $\Sigma^1_1$ (although I could easily be missing something obvious).

One quick observation is that $\mathsf{PA}_{\Sigma^1_1}$ does entail true first-order arithmetic. Given a first-order formula $\varphi(x)$, let $\hat{\varphi}(x)$ be the $\Sigma^1_1$ formula "There is a cut containing $x$ such that every element of the cut satisfies $\varphi$." If $M\models\mathsf{PA}_{\Sigma^1_1}$ we trivially have $\hat{\varphi}^M\in\{\emptyset,M\}$; by induction on the complexity of $\varphi$ we can show that if every standard natural number satisfies $\varphi$ then $0\in\hat{\varphi}^M$ and consequently $M\models\forall x\varphi(x)$ (which then gives $M\equiv\mathbb{N}$). However, I don't see how to use this to get categoricity. In fact, as far as I know it's possible that e.g. every nontrivial ultrapower of $\mathbb{N}$ satisfies $\mathsf{PA}_{\Sigma^1_1}$. (Note that $\Sigma^1_1$ sentences are preserved under taking ultrapowers; however, an instance of induction for a $\Sigma^1_1$ formula is $\Sigma^1_1\vee\Pi^1_1$ and $\Pi^1_1$ sentences are not preserved under taking ultrapowers, so this doesn't seem to help.)


1 Answer 1


If you allow your $\Sigma^1_1$ formulas to have parameters, then PA$_{\Sigma^1_1}$ has only the standard model. To prove it, use the $\Pi^1_1$ definition of standardness to produce a $\Sigma^1_1$ formula $\sigma(x,y)$ saying that $x<y$ and $y-x$ is not standard, i.e., $x$ is infinitely far below $y$. It's easy to show that $\sigma(x,y)$ implies $\sigma(x+1,y)$. So, by $\Sigma^1_1$ induction, if $\sigma(0,y)$ then $\forall x\,\sigma(x,y)$ and, in particular, $\sigma(y,y)$, which is absurd. So $\neg\sigma(0,y)$. But that means $y$ is standard.

  • $\begingroup$ Quite nice, I feel silly now! I am now interested in the parameter-free case, but I definitely had in mind the with-parameters case so this answers my question. $\endgroup$ Feb 22, 2021 at 21:16
  • $\begingroup$ Having no restraint I've now asked an obvious follow-up question. $\endgroup$ Feb 22, 2021 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.