Open Set minus Closed Set is Open

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

For $A \subseteq S$ denote by $\complement_S \left({A}\right)$ the relative complement of $A$ in $S$.

Let $U \in \tau$ and $\complement_S \left({V}\right) \in \tau$.


Then:

$U \setminus V \in \tau$

and:

$\complement_S \left({V \setminus U}\right) \in \tau$


Proof

From Set Difference as Intersection with Relative Complement:

$U \setminus V = U \cap \complement_S \left({V}\right)$

Since $\tau$ is a topology:

$U, \complement_S \left({V}\right) \in \tau \implies U \cap \complement_S \left({V}\right) \in \tau \implies U \setminus V \in \tau$

The other statement follows mutatis mutandis.

$\blacksquare$


Sources