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Suppose I want to show that $A$ is $\Sigma_8$-complete. By definition, this requires to prove, in particular, that $A$ is itself a $\Sigma_8$ set. But I don't really understand if I need to prove as well that $\Sigma_8$ is the "best" one could get? (I.e., that $A$ is not $\Sigma_i$ or $\Pi_i$ for $i=1,2,3,4,5,6,7$.) Some $\Sigma_8$ sets may be as well $\Sigma_1$, but they still qualify for being $\Sigma_8$...

user557
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There are three different notions here:

  • Being $\Sigma_8$.

  • Being properly $\Sigma_8$, which is to say being $\Sigma_8$ but not $\Sigma_n$ for any $n<8$ (or $\Pi_k$ either for that matter for any $k\le 8$).

  • Being $\Sigma_8$-complete.

If we want to show that something is $\Sigma_8$-complete, it is - as you say - not enough to simply show that it is $\Sigma_8$. But it is also not enough to show that it is properly $\Sigma_8$! $\Sigma_8$-completeness is a very strong property: we have to show that any other $\Sigma_8$ set is many-one reducible to the given set, and that doesn't follow from proper-$\Sigma_8$-ness alone. In order to prove that a set $A$ is $\Sigma_8$ complete, we have to prove two things:

  • $A$ is $\Sigma_8$.

  • Every $\Sigma_8$ set $B$ can be many-one reduced to $A$.

That's it - we never need to refer to proper-$\Sigma_8$-ness, and we don't get any benefit from doing so either.

(That said, as a matter of practice basically every naturally-occurring properly-$\Sigma_n$ set is $\Sigma_n$-complete. So you can always take a proof of proper-$\Sigma_n$-ness as really good evidence for $\Sigma_n$-completeness. But this isn't a theorem or anything, it's just a fact about the sorts of sets we run into in mathematical practice ... at least, so far.)

Note, incidentally, that it is not immediately obvious that $\Sigma_8$-completeness implies properly-$\Sigma_8$-ness! To get that you need to use the fact that the arithmetical hierarchy doesn't collapse, which takes a proof. Of course that's not hard, but it is worth noting - and it's worth keeping in mind that there are hierarchies out there which do, or may as far as we know, collapse.

Noah Schweber
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  • Now it's clear. I wonder why textbooks (e.g. Soare's textbook) don't distinguish between being $\Sigma_n$ and being properly $\Sigma_n$. How is the reader supposed to know that the definition of $\Sigma_n$ completeness talks about being properly $\Sigma_n$?.. As for the second requirement in the definition of $\Sigma_n$ completeness, I intentionally didn't mention it (that's why I used "in particular" in my question) since my question is about the first requirement. – user557 Feb 21 '21 at 03:55
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    @user634426 "How is the reader supposed to know that the definition of $\Sigma_n$ completeness talks about being properly $\Sigma_n$?" Huh? It doesn't talk about it at all. $\Sigma_n$-completeness is defined purely in terms of $\Sigma_n$-ness, there's no need to mention proper-$\Sigma_n$-ness at all. I don't see what's missing in e.g. Soare, here. – Noah Schweber Feb 21 '21 at 03:56
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    @user634426 I've edited to, I hope, clarify this point. – Noah Schweber Feb 21 '21 at 03:58
  • Ohh, I thought to show that $A$ is $\Sigma_n$ complete, one has to show that it's properly $\Sigma_n$ and every $\Sigma_n$ set can be $m$-reduced to it. But now I see that one doesn't need to show proper $\Sigma_n$-ness. – user557 Feb 21 '21 at 04:01
  • Does a proof of proper-$\Sigma_n$-ness also a "really good evidence" for $\Sigma_n$-completeness if $n$ is other than $8$? – user557 Feb 21 '21 at 04:06
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    @user634426 Yes, that's for all $n$. I've fixed to clarify that. – Noah Schweber Feb 21 '21 at 04:12
  • I posted a related question here to avoid posting too many comments here. – user557 Feb 21 '21 at 04:45