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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $5$