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 $F_{\sigma\delta}$?

## 1 Answer

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 $X$. $\quad\square$