Uncountable Fort Space is not First-Countable
Theorem
Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a first-countable space.
Proof
Let $\UU$ be a countable set of open neighborhoods of $p$.
Let $U \in \UU$.
Then as $p \in U$, $p \notin \relcomp S U$ and so for $U$ to be open it must follow that $\relcomp S U$ is finite.
From De Morgan's Laws: Complement of Intersection we have:
- $\ds H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S {\bigcap \UU}$
From Countable Union of Countable Sets is Countable it follows that $H$ can be no more than countable.
So $\ds \bigcap \UU = \relcomp S H$ must be uncountable.
So (trivially):
- $H \ne \set p$
So:
- $\exists q \ne p: q \in \ds \bigcap \UU$
So $\relcomp S {\set q}$ is an open neighborhood of $P$ which does not contain any of the elements of $\UU$.
So $\UU$ is not a countable local basis of $p$.
Hence by definition $T$ is not a first-countable space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $24$. Uncountable Fort Space: $5$