# Uncountable Finite Complement Space is not First-Countable

## Theorem

Let $T = \left({S, \tau}\right)$ 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:

- $\forall B \in \mathcal B_x: x \in B$

and such that:

- every open neighborhood of $x$ contains some $B \in \mathcal B_x$.

So:

\(\displaystyle \bigcap \mathcal B_x\) | \(=\) | \(\displaystyle \left\{ {x}\right\}\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle S \setminus \left\{ {x}\right\}\) | \(=\) | \(\displaystyle S \setminus \bigcap \mathcal B_x\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \bigcup_{B \mathop \in \mathcal B_x} \left({S \setminus B}\right)\) | 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 $\displaystyle \bigcup_{B \mathop \in \mathcal B_x} \left({S \setminus B}\right)$ is countable.

So $S \setminus \left\{{x}\right\}$ 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

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 19: \ 4$