Is $\Sigma^{0}_n$ closed under intersection? I think yes m I correct?
Let $L_1$ be an $\Sigma^{0}_1$ complete language, and $L_2$ be a $\Pi^{0}_1$ complete language, such that $\emptyset \neq L= L_1\cap L_2$. Place the language $L$ in the arithmetic hierarchy.
will the language L will be in $\Delta^{0}_2$ am i correct if not pls tell me why. Thanks
$\Sigma_{n}^0$ is not closed under intersection. Note that countable intersection amounts to a $(\forall)$ quantifier. So it suffices to produce a $\Pi_{n + 1}^0$ set which is not $\Sigma_n^0$. Below is a concrete example for $n = 1$.
For example for each $m$, $U_m = \{e : (\exists k > m)\Phi_e(k)\downarrow\}$ is a $\Sigma_1^0$ subset of $\omega$, where $\Phi_e$ denote the $e^\text{th}$ Turing program.
Then $\text{Inf} = \{e : \Phi_e(k)\downarrow \text{ for infinitely many } k\} = \bigcap_{m} U_m$. $\text{Inf}$ is well-known to be $\Pi_2^0$ complete. Hence this set is a intersection of $\Sigma_1^0$ sets which is not $\Sigma_1^0$. (See textbook by Soare for more details.)
The existence of universal $\Sigma_n^0$ and $\Pi_n^0$ sets can also by used to show that each level of the hierarchy is distinct.
You can easily show that $\Sigma_{n}^0 \subset \Delta_{n + 1}^0$ and $\Pi_{n}^0 \subset \Delta_{n + 1}^0$. Hence $L_1 \cap L_2 \in \Delta_{n + 1}^0$. However, not all sets in $\Delta_{n + 1}^0$ can be written as such a intersection. For example, for $n = 1$, such sets are called $2$-c.e. By a very easy diagonalization, it is easy to show that there are $\Delta_2^0$ sets which are not 2-c.e. if you use the limit computable characterization of $\Delta_2^0$. (In fact, there are 3-c.e. sets that are not 2-c.e.) (Again see the textbook of Soare for the relevant definitions.)