Limit Points of Infinite Subset of Finite Complement Space

Theorem
Let $T = \left({S, \tau}\right)$ 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 $T$.

Then $\complement_S \left({U}\right)$ is finite.

Suppose $H \cap U = \varnothing$.

Then from Intersection of Complement with Subset is Empty it follows that $H \subseteq \complement_S \left({U}\right)$ and so $H$ is finite.

So if $H$ is infinite it is bound to have a non-empty intersection with every open set in $T$.

Similarly it can be shown that $U \cap H$ is infinite iff $H$ 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 $H$.