Finite Complement Topology is Topology

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Theorem

Let $T = \struct {S, \tau}$ be a finite complement space.


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


Proof

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

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

$\Box$


Let $A, B \in \tau$.

Let $H = A \cap B$.

Then:

\(\ds H\) \(=\) \(\ds A \cap B\)
\(\ds \leadsto \ \ \) \(\ds \relcomp S H\) \(=\) \(\ds \relcomp S {A \cap B}\)
\(\ds \) \(=\) \(\ds \relcomp S A \cup \relcomp S B\) De Morgan's laws: Complement of Intersection

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

Hence their union is also finite.

Thus $\relcomp S H$ is finite.

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

$\Box$


Let $\UU \subseteq \tau$.

From De Morgan's laws: Complement of Union:

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


But as:

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

each of the $\relcomp S U$ is finite.

Hence so is their intersection.

So $\ds \relcomp S {\bigcup \UU}$ is finite.

So by definition:

$\ds \bigcup \UU \in \tau$


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

$\blacksquare$


Sources

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