Limit Points in Fort Space

Theorem
Let $T = \left({S, \tau_p}\right)$ be a Fort space.

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

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

Proof
Suppose $y \in S, y \ne p$.

We have by definition of Fort space that $\left\{{y}\right\}$ is open in $T$.

So there is no $z \in \left\{{y}\right\}: z \ne y, z \in U$.

Hence $y$ can not be a limit point of $U$.