Infinite Subset of Finite Complement Space Intersects Open Sets

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Theorem

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.

Let $H \subseteq S$ be an infinite subset of $S$.


Then the intersection of $H$ with any non-empty open set of $T$ is infinite.


Proof

Let $U \in \tau$ be any non-empty open set of $T$.

Then $\relcomp S U$ is finite.

We have that:

$H = H \cap \paren {U \cup \relcomp S U} = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$

Aiming for a contradiction, suppose $H \cap U$ is finite.

Since $H \cap \relcomp S U \subseteq \relcomp S U$, $H \cap \relcomp S U$ is also finite.

$H = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$ is the union of two finite sets, and so it is finite.

It is a contradiction that $H$ is infinite and finite at the same time, so $H \cap U$ must be infinite.

$\blacksquare$


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