# Countable Complement Space is not Countably Compact

## Theorem

Let $T = \left({S, \tau}\right)$ be a countable complement topology on an uncountable set $S$.

Then $T$ is not a countably compact space.

## Proof

Consider $U \in \tau$.

By definition of countable complement topology, $\complement_S \left({U}\right)$ is countably infinite.

Then for each $x \in \complement_S \left({U}\right)$, $\complement_S \left({U}\right) \setminus \left\{ {x}\right\}$ is countably infinite.

Then for each $x\in \complement_S \left({U}\right)$, $\complement_S \left({\complement_S(U) \setminus \left\{ {x}\right\} }\right) = U \cup \left\{ {x}\right\}$ is open.

Hence $\left\{ {U \cup \left\{ {x}\right\}: x \in \complement_S \left({U}\right)}\right\}$ is an open cover.

This cover is countable because it is equivalent to $\complement_S \left({U}\right)$.

Aiming for a contradiction, suppose $T$ is a countably compact space.

Then by definition there exists a finite subcover $\left\{ {U \cup \left\{ {x_1}\right\}, \ldots, U \cup \left\{ {x_n}\right\} }\right\}$ of $T$.

Then:

$\left({U \cup \left\{ {x_1}\right\} }\right) \cup \ldots \cup \left({U \cup \left\{ {x_n}\right\} }\right) = U \cup \left\{ {x_1, \ldots, x_n}\right\}$

Since $\complement_S \left({U}\right)$ is not finite, it follows that $U \cup \left\{ {x_1, \ldots, x_n}\right\} \ne S$.

Thus there is no finite subcover of $T$.

It follows by Proof by Contradiction that $T$ cannot be a countably compact space.

$\blacksquare$