Limit Points in Fort Space

Theorem
Let $T = \struct {S, \tau_p}$ be a Fort space.

Let $x \in S$ such that $x \ne p$.

Then $p$ is the only limit point of $x$.

Counterargument
From, we have $\relcomp S {\set x} \in \tau_p$.

For any $y \ne x$, $y \in \relcomp S {\set x}$.

Therefore $\relcomp s {\set x}$ is an open neighborhood of $y$.

From we also have $x \notin \relcomp S {\set x}$.

Hence $y$ is not a limit point of $x$.

By, $x$ cannot be a limit point of $x$.

Therefore $x$ has no limit points in $S$.