Finite Complement Topology is Topology

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:

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:

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

$\Box$

$\tau$ obeys all of the open set axioms, so it is a topology on $T$.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Example $6$
• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.7$
• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology