# Uncountable Finite Complement Space is not First-Countable

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## Theorem

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

Then $T$ is not first-countable.

## Proof

Aiming for a contradiction, suppose some $x \in S$ has a countable local basis.

That means:

there exists a countable set $\BB_x \subseteq \tau$

such that:

$\forall B \in \BB_x: x \in B$

and such that:

every open neighborhood of $x$ contains some $B \in \BB_x$.

Let $y \in S$ with $y \ne x$.

Because its complement relative to $S$ is finite, we have that $S \setminus \set y$ is an open set containing $x$.

Because $\BB_x$ is a neighborhood basis at $x$, there exists some $B \in \BB_x$ such that:

$B \subseteq S \setminus \set y$

whence $y \notin B$.

So:

$y \notin \bigcap \BB_x$

So:

 $\ds \bigcap \BB_x$ $=$ $\ds \set x$ $\ds \leadsto \ \$ $\ds S \setminus \set x$ $=$ $\ds S \setminus \bigcap \BB_x$ $\ds$ $=$ $\ds \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ De Morgan's Laws: Difference with Intersection

By definition, each of $S \setminus B$ is finite.

From Countable Union of Countable Sets is Countable it follows that $\displaystyle \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is countable.

So $S \setminus \set x$ and therefore $S$ is also countable.

From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false.

Hence by definition $T$ can not be first-countable.

$\blacksquare$