Underlying Set of Topological Space is Closed

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

Then the underlying set $S$ of $T$ is closed in $T$.

Proof
From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ $S \setminus U$ is closed in $T$.

From Empty Set is Element of Topology, $\O$ is open in $T$.

From Set Difference with Empty Set is Self:
 * $S \setminus \O = S$

Hence $S$ is closed in $T$.