# Limit Points in Excluded Point Space

## Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ 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 $U \subseteq S$, and let $x \in S, x \ne p$.

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

We have by definition of excluded point 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 $U$.

Similarly $x \notin \left\{{y}\right\}$.

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

Hence the result.

$\blacksquare$