Uncountable Finite Complement Space is not First-Countable
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:
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 $\ds \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ 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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $4$