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
- Mizar article FRECHET:4
- Mizar article YELLOW_8:20