F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space

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Theorem

Let $T = \left({S, \tau}\right)$ be a finite complement topology on an uncountable set $S$.


Then countably infinite subsets of $S$ are $F_\sigma$ sets and are neither open nor closed sets.

Their relative complements in $S$ are $G_\delta$ sets, and are also neither open nor closed sets.


Proof

Let $U$ be a countably infinite subset of $S$.

Then it can be written as:

$\displaystyle U = \bigcup_{x \mathop \in U} \left\{{x}\right\}$

But as $\left\{{x}\right\}$ is a finite subset of $S$, it is by definition the complement of an open set in $T$.

So by definition $\left\{{x}\right\}$ is closed in $T$.


So:

$\displaystyle U = \bigcup_{x \mathop \in U} \left\{{x}\right\}$

is a countable union of closed sets of $T$.

That is, by definition, $U$ is an $F_\sigma$ set.


As $U$ is countably infinite, its complement is uncountable, not finite, and so $U$ is not open in $T$.

But as $U$ is also not finite, it is not the complement of an open set and so is not closed in $T$ either.


From Complement of $F_\sigma$ Set is $G_\delta$ Set it follows that the complement of $U$ is a $G_\delta$ set, and similarly neither open nor closed.

$\blacksquare$


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