Result on D249: F-sigma set
Complement of a G-delta set is an F-sigma set

Let $T = (X, \mathcal{T})$ be a D1106: Topological space such that
 (i) $E \subseteq X$ is a D248: G-delta set in $T$ (ii) $X \setminus E$ is the D79: Set complement of $E$ within $X$
Then $X \setminus E$ is a D249: F-sigma set in $T$.
Proofs

Let $T = (X, \mathcal{T})$ be a D1106: Topological space such that
 (i) $E \subseteq X$ is a D248: G-delta set in $T$ (ii) $X \setminus E$ is the D79: Set complement of $E$ within $X$
Since $E$ is a G-delta set, there are open sets $U_0, U_1, U_2, \ldots \in \mathcal{T}$ such that $$E = \bigcap_{n \in \mathbb{N}} U_n$$ By definition, $X \setminus U_n$ is now a closed set in $T$ for each $n \in \mathbb{N}$. Applying result R4168: Difference of set and countable intersection equals union of differences, one has $$X \setminus E = X \setminus \bigcap_{n \in \mathbb{N}} U_n = \bigcup_{n \in \mathbb{N}} (X \setminus U_n)$$ We have found an expression for $E$ as a countable union of closed sets, so the proof is complete. $\square$