Partition Topology is not Completely Hausdorff

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 not a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Proof

Aiming for a contradiction, suppose $T$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Then from Completely Hausdorff Space is Hausdorff Space, $T$ is a $T_2$ (Hausdorff) space.

This contradicts the result Partition Topology is not Hausdorff.

Hence the result, by Proof by Contradiction.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $5$. Partition Topology: $2$