Open Set minus Closed Set is Open

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Theorem

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

For $A \subseteq S$ denote by $\relcomp S A$ the relative complement of $A$ in $S$.

Let $U \in \tau$ and $\relcomp S V \in \tau$.


Then:

$U \setminus V \in \tau$

and:

$\relcomp S {V \setminus U} \in \tau$


Proof

From Set Difference as Intersection with Relative Complement:

$U \setminus V = U \cap \relcomp S V$

Since $\tau$ is a topology:

$U, \relcomp S V \in \tau \implies U \cap \relcomp S V \in \tau \implies U \setminus V \in \tau$

The other statement follows mutatis mutandis.

$\blacksquare$


Sources