Uncountable Fort Space is not First-Countable

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Theorem

Let $T = \left({S, \tau_p}\right)$ be a Fort space on an uncountable set $S$.


Then $T$ is not a first-countable space.


Proof

Let $\mathcal U$ be a countable set of open neighborhoods of $p$.

Let $U \in \mathcal U$.

Then as $p \in U$, $p \notin \complement_S \left({U}\right)$ and so for $U$ to be open it must follow that $\complement_S \left({U}\right)$ is finite.

From De Morgan's Laws: Complement of Intersection we have:

$\displaystyle H := \bigcup_{U \mathop \in \mathcal U} \complement_S \left({U}\right) = \complement_S \left({\bigcap \mathcal U}\right)$

From Countable Union of Countable Sets is Countable it follows that $H$ can be no more than countable.

So $\displaystyle \bigcap \mathcal U = \complement_S \left({H}\right)$ must be uncountable.

So (trivially):

$H \ne \left\{{p}\right\}$

So:

$\exists q \ne p: q \in \displaystyle \bigcap \mathcal U$

So $\complement_S \left({\left\{{q}\right\}}\right)$ is an open neighborhood of $P$ which does not contain any of the elements of $\mathcal U$.

So $\mathcal U$ is not a countable local basis of $p$.

Hence by definition $T$ is not a first-countable space.

$\blacksquare$


Sources