Complement of F-Sigma Set is G-Delta Set

Theorem
Let $\left({X, \vartheta}\right)$ be a topological space.

Let $S$ be an $F_\sigma$ set of $X$.

Then its complement $X \setminus S$ is a $G_\delta$ set of $X$.

Similarly, let $T$ be an $G_\delta$ set of $X$.

Then its complement $X \setminus T$ is an $F_\sigma$ set of $X$.

Proof
Let $S$ be an $F_\sigma$ set of $X$.

Then $S = \bigcup \Bbb S$ where $\Bbb S$ is a countable union of closed sets in $X$.

Then by De Morgan's Laws (Set Theory) we have:
 * $\displaystyle X \setminus \bigcup \Bbb S = \bigcap_{S \in \Bbb S} \left({X \setminus S}\right)$

By definition of closed set, each of the $X \setminus S$ are open sets.

So $\displaystyle \bigcap_{S \in \Bbb S} \left({X \setminus S}\right)$ is a countable intersection of open sets in $X$.

Hence it is therefore, by definition, a $G_\delta$ set of $X$.

Let $T$ be a $G_\delta$ set of $X$.

Let $T = \bigcap \Bbb T$ where $\Bbb T$ is a countable intersection of open sets in $X$.

Then by De Morgan's Laws (Set Theory) we have:
 * $\displaystyle X \setminus \bigcap \Bbb T = \bigcup_{T \in \Bbb T} \left({X \setminus T}\right)$

By definition of closed set, each of the $X \setminus T$ are closed sets.

So $\displaystyle \bigcup_{T \in \Bbb T} \left({X \setminus T}\right)$ is a countable union of closed sets in $X$.

Hence it is therefore, by definition, an $F_\sigma$ set of $X$.