4
$\begingroup$

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 $F_{\sigma\delta}$?

$\endgroup$
2
$\begingroup$

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 $X$. $\quad\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.