# The Borel class of a countable union of $G_\delta$-sets, which are absolute $F_{\sigma\delta}$

Problem. Assume that a metrizable separable space $$X$$ is the countable union $$X=\bigcup_{n\in\omega}X_n$$ of pairwise disjoint $$G_\delta$$-sets $$X_n$$ in $$X$$ such that each $$X_n$$ is an absolute $$F_{\sigma\delta}$$-set. Is $$X$$ an absolute $$F_{\sigma\delta}$$?

Theorem. Each $$G_{\delta\sigma}$$-subset $$A$$ of a Polish space $$X$$ can be written as the union $$\bigcup_{n\in\omega}A_n$$ of a sequence $$(A_n)_{n\in\omega}$$ of pairwise disjoint $$G_\delta$$-sets in $$X$$.
Proof. Write the set $$A$$ as the union $$A=\bigcup_{n\in\omega}G_n$$ of $$G_\delta$$-sets $$G_n$$ in $$X$$ such that $$\emptyset=G_0\subseteq G_n\subseteq G_{n+1}$$ for all $$n$$. For every $$n\in\omega$$ write the $$G_\delta$$-set $$G_n$$ as the intersection $$G_n=\bigcap_{m\in\omega}U_{n,m}$$ of open sets $$U_{n,m}$$ such that $$U_{n,m+1}\subseteq U_{n,m}\subseteq U_{n,0}=X$$ for all $$m$$. Observe that $$G_{n+1}\setminus G_n=\bigcup_{m\in\omega}(G_{n+1}\cap U_{n,m}\setminus U_{n,m+1})$$ and each set $$G_{n+1}\cap U_{n,m}\setminus U_{n,m+1}$$ is of type $$G_\delta$$ in $$X$$. Now we see that the $$G_{\delta\sigma}$$-set $$A$$ is the union $$A=\bigcup_{n\in\omega}G_{n+1}\setminus G_n=\bigcup_{n\in\omega}(G_{n+1}\cap U_{n,m}\setminus U_{n,m+1})$$of the countable family $$\big(G_{n+1}\cap U_{n,m}\setminus U_{n,m+1}\big)_{n,m\in\omega}$$of pairwise disjoint $$G_\delta$$-subsets of $$X$$. $$\quad\square$$