Limit Points of Infinite Subset of Finite Complement Space
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Theorem
Let $T = \struct {S, \tau}$ be a finite complement space.
Let $H \subseteq S$ be an infinite subset of $S$.
Then every point of $S$ is a limit point of $H$.
Proof
Let $U \in \tau$ be any open set of $T$.
From Infinite Subset of Finite Complement Space Intersects Open Sets, we have that $U \cap H$ is infinite if and only if $H$ is infinite.
Let $x \in S$.
Then every open set $U$ in $T$ such that $x \in U$ also contains an infinite number of points of $H$ other than $x$.
Thus, by definition, $x$ is a limit point of $H$.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $1$