Partition Topology is T5

Theorem
Let $S$ be a set and let $\PP$ be a partition on $S$ which is not the (trivial) partition of singletons.

Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.

Then $T$ is a $T_5$ space.

Proof
Let $A$ and $B$ be subsets of $S$ such that $A^- \cap B = A \cap B^- = \O$.

From Open Set in Partition Topology is also Closed, we get that $A^-$ and $B^-$ are both clopen.

From Set is Subset of its Topological Closure:
 * $A \subseteq A^-$

From $A^- \cap B = \O$ it follows from Intersection with Complement is Empty iff Subset that $B \subseteq S \setminus \paren {A^-}$.

Since $A^-$ is clopen, it follows that $A^-$ and $S \setminus \paren {A^-}$ are both open.

Also, they are disjoint by definition.

Thus we have $A^-$, an open set containing $A$, and $S \setminus \paren {A^-}$, an open set containing $B$.

Thus, by definition, $T$ is a $T_5$ space.