I have started reading about Borel Hierarchies. If I understand correctly, Then $\Sigma_2^0$ is the collection of all sets of the form $A=\bigcup_{n \in \omega}{B_n}^c$, where $B_n$ is open. also, $\Sigma_3^0$ is the collection of all sets of the form $A=\bigcup_{n \in \omega}{B_n}^c$, where $B_n$ is in $\Sigma_{2}^0$ or in $\Sigma_{1}^0$. I can think of a set which is in $\Sigma_{3}^0$ and not in $\Sigma_{2}^0$ or in $\Sigma_{1}^0$, for example, the set, $(-\infty,0] \cup \bigcup_{n \in \mathbb{N}} (\frac{1}{n+1},\frac{1}{n}) \cup (1,\infty)$. But, is there an example of a set which is in $\Sigma_{4}^0$ but not in $\Sigma_{3}^0$ or in $\Sigma_{2}^0$ or in $\Sigma_{1}^0$?
Thank you! Shir
You cannot construct a $\Sigma_4^0$ Borel set, for one major reason, it is consistent with the failure of choice that every set is $\Sigma_4^0$.
The main method of showing that there are such sets is to prove the existence of $\Sigma^0_n$-universal sets. You can read my question, and answer, about those here.