Partition Topology is not Completely Hausdorff

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Theorem

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

Let $T = \left({S, \tau}\right)$ be the partition space whose basis is $\mathcal P$.


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


Proof

Aiming for a contradiction, suppose $T$ is a $T_2$ (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.

$\blacksquare$


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