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> **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 $F\_{\sigma\delta}$?
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The Borel class of a countable union of $G_\delta$-sets, which are absolute $F_{\sigma\delta}$
The answer to this question is negative and follows from


**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 $X$. $\quad\square$


