Higher-Aleph Complement Topology is Topology

Theorem

Let $T = \struct {S, \tau}$ be an $\aleph_n$ complement space.

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

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

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

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 $

$=$

$\ds \relcomp S A \cup \relcomp S B$

But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ both have a cardinality strictly smaller than that of $S$.

Hence their union also has a cardinality strictly smaller than that of $S$.

So $\relcomp S H$ has a cardinality strictly smaller than that of $S$.

So $H = A \cap B \in \tau$ as its complement has a cardinality strictly smaller than that of $S$.

Now let $\UU \subseteq \tau$.

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

But as:

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

each of the $\relcomp S U$ has a cardinality strictly smaller than that of $S$.

Hence so has their intersection.

So $\ds \relcomp S {\bigcup \UU}$ has a cardinality strictly smaller than that of $S$, which means:

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

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

$\blacksquare$