Complement of Closed Set is Open Set

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Theorem

Let $T = \struct{S, \tau}$ be a topological space.

Let $F \subseteq S$ be a closed set in $T$.

Then:

$S \setminus F \in \tau$

Proof

By definition of closed set:

$\exists U \in \tau : F = S \setminus U$

From Relative Complement inverts Subsets of Relative Complement:

$U = S \setminus F$

The result follows.

$\blacksquare$