Fortissimo Space is not First-Countable

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \struct {S, \tau_p}$ be a Fortissimo space.


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


Proof

This proof follows the proof from Countable Complement Space is not First-Countable.


Aiming for a contradiction, suppose that $p \in S$ has a countable local basis.

That means:

there exists a countable set of sets $\BB_p \subseteq \tau$

such that:

$\forall B \in \BB_p: p \in B$

and such that:

every open neighborhood of $p$ contains some $B \in \BB_p$.


So:

\(\ds \bigcap \BB_p\) \(=\) \(\ds \set p\)
\(\ds \leadsto \ \ \) \(\ds S \setminus \set p\) \(=\) \(\ds S \setminus \bigcap \BB_p\)
\(\ds \) \(=\) \(\ds \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}\) De Morgan's Laws: Difference with Intersection

By definition of the fortissimo space, each of $S \setminus B$ is countable.

From Countable Union of Countable Sets is Countable it follows that $\ds \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}$ is also countable.

So $S \setminus \set p$ and therefore $S$ is also countable.

From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $p \in S$ has a countable local basis must be false.

Hence by definition $T$ can not be first-countable.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Fortissimo_Space_is_not_First-Countable&oldid=545466"