Countable Complement Topology is Topology
Theorem
Let $T = \struct {S, \tau}$ be a countable 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 countable.
Now suppose $A, B \in \tau$.
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 countable.
Hence from Countable Union of Countable Sets is Countable their union is also countable and so $\relcomp S H$ is countable.
So $H = A \cap B \in \tau$ as its complement is countable.
Now let $\UU \subseteq \tau$.
Then:
- $\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
by De Morgan's Laws: Complement of Union.
But as:
- $\forall U \in \UU: \relcomp S U \in \tau$
each of the $\relcomp S U$ is countable.
Hence so is their intersection.
So $\ds \relcomp S {\bigcup \UU}$ is countable which means:
- $\ds \bigcup \UU \in \tau$
So $\tau$ is a topology on $T$.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology