Open Set in Partition Topology is also Closed

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

Then $T$ is a partition space :
 * $\forall U \subseteq S: U \in \tau \iff \relcomp S U \in \tau$

That is, a topological space is a partition space all open sets are closed, and all closed sets are also open.

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

Necessary Condition
Let $T$ be a partition space.

Then there exists a partition $\PP$ which forms the basis for $T$.

Let $U \in \tau$.

Then $U$ is the union of elements of $\PP$.

Then by definition $\relcomp S U$ is closed in $T$.

Then $\relcomp S U$ is the union of all the other elements of $\PP$.

That is, $\relcomp S U$ is also open in $T$.

By definition, then, $\relcomp S {\relcomp S U} = U$ is closed in $T$.

Sufficient Condition
Now suppose that every open set in $T$ is closed, and every closed set in $T$ is open.

$T$ is not a partition space.

Thus, by definition, there exists an open set of $T$ which is not the union of disjoint sets.