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

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

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

So $$\bigcap_{S \in \mathbb S} \left({X \setminus S}\right)$$ 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 $$T = \bigcap \mathbb T$$ where $$\mathbb T$$ is a countable intersection of open sets in $$X$$.

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

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

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

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