Limit Points in Excluded Point Space

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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.

$\blacksquare$


Sources