Limit Points in Uncountable Fort Space

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Theorem

Let $T = \struct {S, \tau_p}$ be an uncountable Fort space.

Let $U \subseteq S$ be a countably infinite subset of $S$.


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


Proof

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

We have by definition of Fort 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 $U$.


Suppose $p \in V \in \tau_p$ for some $V \subseteq S$.

From Definition of Fort Space, $\relcomp S V$ is finite.

Then:

\(\displaystyle \relcomp S {V \setminus \set p}\) \(=\) \(\displaystyle \paren {\relcomp S V} \cup \paren {V \cap \set p}\) Set Difference with Set Difference is Union of Set Difference with Intersection
\(\displaystyle \) \(\subseteq\) \(\displaystyle \paren {\relcomp S V} \cup \set p\) Set Union Preserves Subsets: $V \cap \set p \subset \set p$: Intersection is Subset

Thus $\relcomp S {V \setminus \set p}$ is also finite.

Since $U$ is countably infinite:

$U \not \subseteq \relcomp S {V \setminus \set p}$.

From Empty Intersection iff Subset of Complement:

$U \cap \paren {V \setminus \set p} \ne \O$.


Since $V$ is arbitrary, every open neighborhood of $p$ has non-empty intersection with $U$.

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

$\blacksquare$


Sources