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This is a follow-up to this answer.

How does the fact that the arithmetic hierarchy doesn't collapse imply that if $A$ is $\Sigma_n$ complete, then $A$ is properly-$\Sigma_n$?

Some thoughts (but nothing serious). Assume that $A$ is not properly-$\Sigma_n$. Then it's either $\Sigma_i$ for $0\leq i\leq n-1$ or $\Pi_i$ for $0\leq i\leq n$. Suppose the former first. We know that any $\Sigma_n$ set is $m$-reducible to $A$. Should I consider a proper-$\Sigma_n$ set and show somehow that it's not possible to $m$-reduce it to $A$ (if this is even true)? I don't really see how the properness or inclusions in the hierarchy would be used.

Update: here are the definitions.

A set $B$ is $\Sigma_0$ ($\Pi_0$, $\Delta_0$) if $B$ is computable.

For $n\ge 1$, a set $B$ is $\Sigma_n$ if there is a computable relation $R(x,y_1,\dots,y_n)$ such that $$x\in B\iff (\exists y_1)(\forall y_2)\dots (Q y_n) R(x,y_1,\dots, y_n)$$ where $Q$ is $\exists$ for $n$ odd and $\forall$ for $n$ even.

The definition of $\Pi_n$ sets for $n\ge 1$ is similar (just switch the quantifiers).

A set $B$ is $\Delta_n$ if it's both $\Sigma_n$ and $\Pi_n$

A set $B$ is $\Sigma_n$-complete if $B$ is $\Sigma_n$ and for any $\Sigma_n$ set $A$, one has $A\leq_m B$.

user557
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  • Somewhat related to the answer below: https://math.stackexchange.com/questions/4033789/a-leq-m-b-and-b-in-pi-n-implies-a-in-pi-n – user557 Feb 21 '21 at 22:31

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The key fact here is that if a set $A$ qualifies for some level of the hierarchy, then so does any set m-reducible to $A.$ Thus if $A$ is $\Sigma_n$ complete, and also qualifies for some lower level, then any $\Sigma_n$ set also qualifies for that lower level and the hierarchy collapses.

To prove the key fact based on your definition for the levels of the hierarchy, just observe that if $f$ is a total computable function and $R(n,y_1,\ldots, y_n)$ is a computable relation, then the relation $R'$ defined by $$R'(n,y_1,\ldots,y_n)\iff R(f(n),y_1,\ldots,y_n) $$ is computable. Thus if $B$ is m-reducible to $A,$ then there is a total computable $f$ such that $n\in B$ iff $f(n)\in A,$ and we can see from the definition this means that $B$ qualifies for any level that $A$ qualifies for.

spaceisdarkgreen
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  • What is a $\Delta_1$ function? From what I know, $f$ should be a total computable function. And what does it mean for a sentence to be $\Delta_m$? I only know the definition of a set being $\Delta_m$. – user557 Feb 21 '21 at 18:59
  • Also, why is it enough to consider $A\in \Delta_m$? If $A$ is not $\Sigma_n$, it could be $\Sigma_{n-1}$ but not $\Delta_{n-1}$, right? – user557 Feb 21 '21 at 19:01
  • @user634426 A total computable function and a $\Delta_1$ function are the same thing. ($\Delta_1$ function means the defining formula $z = f(n)$ is $\Delta_1.$) I should have said $f(n)\in A$ is a $\Delta_m$ formula in the variable $n$ (i.e. equivalent to both a $\Sigma_m$ and a $\Pi_m$ formula). WRT to the last thing, if it is $\Sigma_{n-1}$ then it is $\Delta_n.$ – spaceisdarkgreen Feb 21 '21 at 20:12
  • Actually, in my original answer, I had started "Say $A$ is $\Sigma_m$ for some $m<n$..." (notice the strict inequality) and changed every $\Delta$ to a $\Sigma$, with no other changes. Then I realized this doesn't include the case where $\Sigma_n$ might collapse to $\Delta_n,$ and not necessarily to $\Sigma_m$ for some $m<n,$ so I edited to the present version. – spaceisdarkgreen Feb 21 '21 at 20:19
  • Sorry, I'm still confused by formulas. I know the notion of $\Delta_n$-ness for sets. Is there a way to rephrase the statement "$f(n)\in A$ is $\Delta_m$" in terms of a set being $\Delta_m$? Plus when "formulas" come into the picture, what's not fully clear to me is the definition of a formula (i.e. what's the syntax from which formulas are constructed and what syntactic operations give formulas)... – user557 Feb 21 '21 at 20:37
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    @user634426 The definition of the arithmetic hierarchy in terms of definability by formulas in the language of first order arithmetic is the standard one from my perspective. Are you using a pure computability theory formulation (e.g. a set is $\Sigma_n$ iff it is enumerable by a $\emptyset^{(n-1)}$-Turing machine)? If you say what definitions you're using I can probably phrase the answer in a way that is more transparently connected to them. – spaceisdarkgreen Feb 21 '21 at 20:40
  • I'm using this definition: a set $A$ is $\Sigma_n$ if $x\in A\iff \exists y_1\forall y_2\dots Qy_n (x,y_1,\dots, y_n)\in R$ for a computable set $R\subset \mathbb N^{n+1}$. – user557 Feb 21 '21 at 20:44
  • Strictly speaking it doesn't make sense to talk about $\Delta_n$ formulas (for $n>0$ anyways) since that's not a syntactic condition. But this language does get used anyways: when a particular theory (resp. structure) is specifically considered, "$\Delta_n$ formula" means "formula which the theory proves is equivalent to a $\Sigma_n$ formula and to a $\Pi_n$ formula" (resp. "formula which defines a $\Delta_n$ subset of the structure in question"). In this case the structure in mind is the standard model of arithmetic. – Noah Schweber Feb 21 '21 at 21:14
  • Thanks for the edit. Now it's more clear what you meant, but I don't see why this solves the problem. What does the fact that $B$ is $\Delta_m$ contradict to? it doesn't contradict that $B$ is $\Sigma_n$, does it? – user557 Feb 21 '21 at 21:30
  • @user634426 $B$ is an arbitrary $\Sigma_n$ set. So this shows if $A$ is at any lower level of the hierarchy, so is any $\Sigma_n$ set... thus the hierarchy collapses. Perhaps also helpful: I phrased the argument for $\Delta$ but the same argument works to show $\Sigma$ and $\Pi$ are also preserved by m-reducibility. The reason I chose to say $\Delta_m$ (for some $m\le n$) was simply because $\Delta_n$ is the next lower-down rung from $\Sigma_n$ ... if I'd said $\Sigma_m$ for some $m<n$ it would skip over this case. – spaceisdarkgreen Feb 21 '21 at 22:01
  • @user634426 I rewrote the answer... hopefully clearer how. – spaceisdarkgreen Feb 21 '21 at 22:14