Particular Point Topology with Three Points is not T4
That is, such that $S$ has more than two distinct elements.
Then $T$ is not a $T_4$ space.
We have that there are at least three elements of $S$.
So, consider $x, y, p \in S: x \ne y, x \ne p, y \ne p$.
Then $X = \set x, Y = \set y$ are closed in $T$ and $X \cap Y = \O$.
Suppose $U, V \in \tau_p$ are open sets in $T$ such that $X \subseteq U, Y \subseteq V$.
But as $p \in U, p \in V$ we have that $U \cap V \ne \O$.
So $T$ is not a $T_4$ space.
Mistakes in Sources
- Every particular point topology is $T_0$, but since there are no disjoint open sets, none of the higher separation axioms are satisfied unless $X$ has only one point.