Bottom in Ordered Set of Topology

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

Let $P = \struct {\tau, \subseteq}$ be an inclusion ordered set of $\tau$.

Then $P$ is bounded below and $\bot_P = \O$

Proof
By Empty Set is Element of Topology:
 * $\O \in \tau$

By Empty Set is Subset of All Sets:
 * $\forall A \in \tau: \O \subseteq A$

Hence $P$ is bounded below.

Thus by definition of the smallest element:
 * $\bot_P = \O$