Is every generalized-$F_{\sigma}$ set an $F_{\sigma}$ set? A subset $S$ of a topological space $X$ is called a generalized-$F_{\sigma}$ set in $X$ if for all open $G \subset X$ with $S\subset G$, there exists an $F_{\sigma}$-set $F$ such that $S\subset F\subset G$.
It's easy to see that every $F_{\sigma}$-set is a generalized-$F_{\sigma}$ set, but I am not sure that every generalized-$F_{\sigma}$ set is a $F_{\sigma}$-set, and I can't find a counterexample.
 A: Let $X=\omega_1$ with the order topology, and let $S$ be a stationary subset of $\omega_1$ that does not contain a closed, unbounded subset of $\omega_1$. It’s an easy consequence of the pressing-down lemma (Fodor’s lemma) that if $U$ is an open nbhd of $S$, there is an $\alpha<\omega_1$ such that $U\supseteq(\alpha,\omega_1)=[\alpha+1,\omega_1)$, and hence $U$ is itself an $F_\sigma$: it’s the union of the closed set $[\alpha+1,\omega_1)$ and the countably many closed singletons $\{\xi\}$ for $\xi\in U\cap[0,\alpha]$. Thus, $S$ is a generalized $F_\sigma$.
$S$ is not an $F_\sigma$, however, since its only closed subsets are bounded, hence countable, and $S$ itself is uncountable.
A: Recall that in a metric space all open sets are Fσ. It follows that every subset $S$ of a metric space is a generalised-Fσ (given an open $G$ including $S$ take the witnessing Fσ-set to be $G$ itself).
So a counterexample would be any subset of a metric space which is not Fσ; e.g., the set of irrational numbers in the real line.
