Limit Points in Excluded Point Space

Theorem
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.

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

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

Similarly, let $U \subseteq S$.

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

Proof
Let $U \subseteq S$.

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

From:
 * Excluded Point Topology is Open Extension Topology of Discrete Topology
 * Limit Points in Open Extension Space

it follows that:
 * $p$ is a limit point of $U$
 * $p$ is a limit point of $x$.

Now suppose $y \in S$ such that $y \ne p$.

We have by definition of excluded point space that $\set y$ is open in $T$.

So there is no $z \in \set y: z \ne y, z \in U$.

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

Similarly $x \notin \set y$.

So $y$ can not be a limit point of $x$.

Hence the result.