Am reading a text about computability theory, and according to the text, at each level $k$ of the arithmetical hierarchy, we have two sets, $\Sigma_k$ and $\Pi_k$, where $\Pi_k$ is defined as:

$$ \Pi_k=co-\Sigma_k $$

So that for $k=0$, we have the class of decidable sets and $\Sigma_0=\Pi_0$, and for $k=1$, we have $\Sigma_1$ as the class of computably enumerable (c.e.) sets and $\Pi_1$ as the class of not computably enumerable sets (not c.e.)....

Let $L(M_e)$ denote the language recognized by Turing Machine $M_e$ with Godel number $e$. I came across the following language $E$, where:


i.e. $E$ is the language of all Turing Machine codes $e$ that are computably enumerable. By a diagonalization argument, it can be shown that $E$ is not c.e. This implies that:

$$ E \in \Pi_1 $$

However, if $E \in \Pi_1$, it means that $E = co-A$, for some $A \in \Sigma_1$, using the definition in the above statement... However, the complement of $E$ is:

$$ \overline{E}=\overline{\{e|L(M_e)=\Sigma^*\}} $$

which (I guess) means that $\overline{E}$ is the language of all Turing Machines $e$ such that on some inputs, $e$ diverges... However, it has been shown that $\overline{E} \equiv_m K^{2}$, i.e. $\overline{E} \equiv_m K^K$, so that, where given two sets $A$ and $B$, we have $A \equiv_m B$ iff $A \leq_m B$ and $B \leq_m A$, and $\leq_m$ refers to a many-to-one reduction:

$$ \overline{E} \equiv_m K^K \in \Sigma_2 $$

Given that $\Sigma_2 \neq \Sigma_1$, it looks like that $\overline{E}$ is not computably enumerable... But doesn't this contradict the definition of $\Pi_1$ which states that the complement of a not c.e. set is c.e. ?

I think am missing something in my understanding of the definitions ...

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    $\begingroup$ Your first mistake is that if $E \not\in \Sigma_k$ then we need to have $E \in \Pi_k$. This is not the case. There are many problems that are, e.g. $\Sigma_2$-hard, meaning that these problems are neither in $\Sigma_1$ nor in $\Pi_1$. $\endgroup$ – Pål GD Jun 3 at 6:43
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    $\begingroup$ Let me be more explicit than @PålGD: it is not true that $\Pi_1$ is the class of sets which are not c.e. The correct statement is: $\Pi_1$ is the class of sets whose complements are c.e. (A similar mistake is sometimes made by topology students who think that a set is closed if it is not open, but that's wrong.) $\endgroup$ – Andrej Bauer Jun 3 at 8:23

For $k$ even, a language $L$ is in $\Pi_k$ if there exists a recursive predicate $R$ such that $$ x \in L \Longleftrightarrow \forall y_1 \exists y_2 \cdots \forall y_{k-1} \exists y_k \, R(x,y_1,\ldots,y_k) $$ The quantifiers alternate between $\forall$ and $\exists$.

When $k$ is odd, the same definition works, but the last quantifier is $\forall$: $$ x \in L \Longleftrightarrow \forall y_1 \exists y_2 \cdots \exists y_{k-1} \forall y_k \, R(x,y_1,\ldots,y_k) $$

For example, the language of all total Turing machines is $\Pi_2$ since $$ x \in \mathsf{TOT} \Longleftrightarrow \forall y \exists z \, \text{"Machine $x$ halts on input $y$ within $z$ steps"} $$

The class $\Sigma_k$ is defined in the same way, with the first quantifier being $\exists$ rather than $\forall$.

If you complement a language in $\Sigma_k$ you get one in $\Pi_k$, and vice versa. This is due to de Morgan's laws for quantifiers, and the fact that the negation of a recursive predicate is also recursive.

For example, the language of non-total Turing machines is $\Sigma_2$ since $$ x \in \mathsf{NTOT} \Longleftrightarrow x \notin \mathsf{TOT} \Longleftrightarrow \exists y \forall z \, \text{"Machine $x$ doesn't halt on input $y$ within $z$ steps"} $$

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  • $\begingroup$ thanks, does this mean that $E$ (as I wrote above) is really in $\Pi_2$ and not $\Pi_1$ as I thought ? $\endgroup$ – Link L Jun 3 at 8:30
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    $\begingroup$ Your $E$ is the same as my $\mathsf{TOT}$, which is known to be $\Pi_2$-complete. In particular, it is not in $\Pi_1$. $\endgroup$ – Yuval Filmus Jun 3 at 8:54

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