Compact Complement Topology is Topology

Theorem

Let $T = \struct {\R, \tau}$ be the compact complement space.

Then $\tau$ is a topology on $T$.

By definition, we have that $\O \in \tau$.

We also have that $\R \in \tau$ as $\relcomp \R \R = \O$ which is trivially compact.

Now suppose $A, B \in \tau$.

Let $H = A \cap B$.

Then:

$\ds H$

$=$

$\ds A \cap B$

$\ds \leadsto \ \ $

$\ds \relcomp \R H$

$=$

$\ds \relcomp \R {A \cap B}$

$\ds $

$=$

$\ds \relcomp \R A \cup \relcomp \R B$

But as $A, B \in \tau$ it follows that $\relcomp \R A$ and $\relcomp \R B$ are both compact.

Hence from Union of Compact Sets is Compact their union is also compact and so $\relcomp \R H$ is compact.

So $H = A \cap B \in \tau$ as its complement is compact.

Now let $\UU \subseteq \tau$.

$\ds \relcomp \R {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp \R U$

But as:

$\forall U \in \UU: \relcomp \R U \in \tau$

each of the $\relcomp \R U$ is compact.

So $\ds \relcomp \R {\bigcup \UU}$ is compact which means $\ds \bigcup \UU \in \tau$.

So $\tau$ is a topology on $T$.

$\blacksquare$