Let $\{E_n\}_{n\ge 1}$ be a countably infinite family of closed subsets of R. Let $A$ be the set of points which belong to infinitely many of the sets $E_n$. Then it is known that $A=\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty} E_m$, and hence $A$ is an $F_{\sigma\delta}$ (that is, a countable intersection of $F_{\sigma}$ sets). Is there an example of sets $\{E_n\}$ such that $A$ is not an $F_{\sigma}$?
Suggestion: Based on the helpful comments (now deleted) of a contributor, here is a suggestion. It uses $(0,1)$ rather than R. Let $P$ be the countable set $P=\{ 1/2, 1/3, 2/3, 1/5, 2/5, 3/5, 4/5, 1/7, \ldots\}$ (and so on through the reciprocals of primes). Let $F_n$ be the open set consisting of the first $n$ terms of $P$ together with $n$ open intervals of length $1/(n+1)^{n+1}$ centred on these $n$ points. Let $E_n$ be the complement of $F_n$ in $(0,1)$. Then I reckon that the set $A$ of points in $(0,1)$ which belong to infinitely many of the closed sets $E_n$ is disjoint from the dense set $P$, but otherwise contains "most" of the points in $(0,1)$ and hence is not an $F_{\sigma}$. Can anyone clinch the argument, possibly by modifying the sets $E_n$ a bit?