# How to prove that these sets are $\mathbf{\Delta}_\alpha^0$.

I have been studying basics of descriptive set theory lately. In the lecture notes I follow (sadly, the notes are written in Czech), there is a lemma (which is used to prove that the functions of first borel class and the functions of first baire class coincide):

Let X be a metric space and $A_n\in\mathbf{\Sigma}_\alpha^0$ (finitely or countably many). Then there exist disjoint $D_n\subseteq A_n$ such that $D_n\in\mathbf{\Sigma}_\alpha^0$ and $\bigcup\limits_n D_n =\bigcup\limits_n A_n$. Moreover, if $\bigcup\limits_n A_n = X$, then $D_n\in\mathbf{\Delta}_\alpha^0$.

I could prove the first part of the lemma but I can't prove the moreover part. I'd like to show that $X\setminus D_n\in\mathbf{\Sigma}_\alpha^0$. Obviously, we have $X\setminus D_n = \bigcup\limits_n A_n\setminus\bigcup\limits_{k=1}^\infty D_n^k$, where $D_n^k\in\mathbf{\Delta}_\alpha^0$ as $D_n\in\mathbf{\Sigma}_\alpha^0$. However, I can't see how to continue.

Thank you for any help!

## 1 Answer

Well, in fact it's trivial. If $\bigcup\limits_n A_n=\bigcup\limits_n D_n= X$, then (as $D_n$ are disjoint) $D_n^C = \bigcup\limits_{k\neq n} D_k\in\mathbf{\Sigma}_\alpha^0$, hence $D_n\in\mathbf{\Delta}_\alpha^0$.

I'm sorry I'd asked such a trivial question but you certainly know it - sometimes the most obvious things are the hardest one.