Empty Set is Element of Topology

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

Then the empty set $\O$ is an open set of $T$.

Proof
for a topology states that:
 * $\ds \forall \AA \subseteq \tau: \bigcup \AA \in \tau$

By Empty Set is Subset of All Sets, we have that $\O \subseteq \tau$.

Hence, by Union of Empty Set, we have:
 * $\ds \O = \bigcup \O \in \tau$