MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recall that a $J$-structure is an amenable structure of the form ($J_{\alpha}^A,B$) where $A$ and $B$ are predicates and $\alpha$ is a limit ordinal. Then if you let $M=J_{\alpha}^A$, there is a surjective function $f:[\alpha]^{\lneq \omega} \rightarrow M$ which is $\Sigma_1^M$. Can we prove that there exists surjective functions $f:[\alpha]^{\lneq \omega} \rightarrow M$ which are $\Sigma_n^M$ for all $n$<$\omega$?

The proof that there is a surjective function $f:[\alpha]^{\lneq \omega} \rightarrow M$ which is $\Sigma_1^M$ uses that $h_M(\alpha)$ is a $\Sigma_1$ elementary substructure of $M$ ($h_M(\alpha)$ the Skolem hull of $\alpha$)

share|cite|improve this question
I don't understand the question. A $\Sigma^M_1$ function is also (trivially) $\Sigma^M_n$ for all larger $n$. So you seem to already have exactly what you asked for --- unless you really want the case $n=0$. – Andreas Blass Aug 21 '10 at 8:58
That is true, I think I got confused by this small thing. Thanks – Carlo Von Schnitzel Aug 21 '10 at 9:01

Removing the question from unanswered queue by putting answer from comment of Andreas Blass into answer...:

A $\Sigma^M_1$ function is also (trivially) $\Sigma^M_n$ for all larger $n$.

share|cite|improve this answer
I don't fully see the purpose here, but when you turn a comment by someone else to an answer the honest thing is to first ask them to do it, and if they don't answer or reject the idea, post it as cw with reference to the comment. Otherwise it looks like a bad attempt to grab reputation and a necromancer badge. – Asaf Karagila Jan 29 '15 at 5:34
@AsafKaragila updated as suggested – Loreno Heer Jan 29 '15 at 11:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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