# Countable Complement Space is not Countably Compact

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## Theorem

Let $T = \struct {S, \tau}$ 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, $\relcomp S U$ is countably infinite.

Then for each $x \in \relcomp S U$, $\relcomp S U \setminus \set x$ is countably infinite.

Then for each $x \in \relcomp S U$, $\relcomp S {\relcomp S U \setminus \set x} = U \cup \set x$ is open.

Hence $\set {U \cup \set x: x \in \relcomp S U}$ is an open cover.

This cover is countable because it is equivalent to $\relcomp S U$.

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

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

Then:

$\paren {U \cup \set {x_1} } \cup \dotsb \cup \paren {U \cup \set {x_n} } = U \cup \set {x_1, \ldots, x_n}$

Since $\relcomp S U$ is not finite, it follows that $U \cup \set {x_1, \ldots, x_n} \ne S$.

Thus there is no finite subcover of $T$.

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

$\blacksquare$