Uncountable Finite Complement Space is not First-Countable

From ProofWiki
Jump to navigation Jump to search

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:

there exists a countable set $\mathcal B_x \subseteq \tau$

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