Excluded Set Topology is not T0

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Theorem

Let $T = \left({S, \tau_{\bar H}}\right)$ 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.

$\blacksquare$


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