# Infinite Subset of Finite Complement Space Intersects Open Sets

## Theorem

Let $T = \left({S, \tau}\right)$ 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 $\complement_S \left({U}\right)$ is finite.

We have that:

$H = H \cap \left({U \cup \complement_S \left({U}\right)}\right) = \left({H \cap U}\right) \cup \left({H \cap \complement_S \left({U}\right)}\right)$

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

Since $H \cap \complement_S \left({U}\right) \subseteq \complement_S \left({U}\right)$, $H \cap \complement_S \left({U}\right)$ is also finite.

$H = \left({H \cap U}\right) \cup \left({H \cap \complement_S \left({U}\right)}\right)$ 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.

$\blacksquare$