Excluded Set Topology is not T0

Theorem

Let $T = \struct {S, \tau_{\bar H} }$ be an excluded set space where $H$ has at least two distinct points.

Then $T$ is not a $T_0$ (Kolmogorov) space.

Proof

Let $x, y \in H$ such that $x \ne y$.

Then $x, y \in S$, but by definition $x$ and $y$ are in no other open sets of $T$.

Hence there is no $U \in \tau_{\bar H}$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$.

Hence the result, by definition of $T_0$ (Kolmogorov) space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $7$